💡Alright, so I have this problem here that I need to prove two things about sequences involving logarithms and exponents. It seems a bit intimidating at first, but I'll try to break it down step by step.First, the problem states that if ( n ) increases, the numbers in the form ( lg left(1+frac{1}{n}right)^{n} ) increase, while those in the form ( lg left(1+frac{1}{n}right)^{n+1} ) decrease. Then, it mentions that the numbers ( b_{n}=left(1+frac{1}{n}right)^{n+1} ) form a decreasing sequence, and any element of this sequence is greater than all elements of the sequence ( a_{n}=left(1+frac{1}{n}right)^{n} ). Finally, I need to show that the function ( left(1+frac{1}{x}right)^{x} ) increases as ( x ) increases.Okay, let's start by understanding what these sequences are. The sequence ( a_n = left(1 + frac{1}{n}right)^n ) is a well-known sequence that converges to Euler's number ( e ) as ( n ) approaches infinity. Similarly, ( b_n = left(1 + frac{1}{n}right)^{n+1} ) also converges to ( e ), but from above, if I recall correctly.So, the problem is asking me to show that ( a_n ) is increasing and ( b_n ) is decreasing, and that every term of ( b_n ) is greater than every term of ( a_n ). Additionally, I need to show that the function ( f(x) = left(1 + frac{1}{x}right)^x ) is increasing for all ( x ).Let me first focus on the sequences ( a_n ) and ( b_n ). To show that ( a_n ) is increasing, I need to show that ( a_{n+1} > a_n ) for all ( n ). Similarly, to show that ( b_n ) is decreasing, I need to show that ( b_{n+1} < b_n ) for all ( n ).Let me write out ( a_{n+1} ) and ( a_n ):( a_n = left(1 + frac{1}{n}right)^n )( a_{n+1} = left(1 + frac{1}{n+1}right)^{n+1} )I need to compare these two. Maybe I can take the ratio ( frac{a_{n+1}}{a_n} ) and show that it's greater than 1.So,( frac{a_{n+1}}{a_n} = frac{left(1 + frac{1}{n+1}right)^{n+1}}{left(1 + frac{1}{n}right)^n} )Hmm, that looks a bit complicated. Maybe taking the logarithm would help, since logarithms turn exponents into multipliers, which might be easier to handle.Let me define ( ln(a_n) = n lnleft(1 + frac{1}{n}right) ) and ( ln(a_{n+1}) = (n+1) lnleft(1 + frac{1}{n+1}right) ).If I can show that ( ln(a_{n+1}) > ln(a_n) ), then ( a_{n+1} > a_n ).So, let's compute the difference:( ln(a_{n+1}) - ln(a_n) = (n+1) lnleft(1 + frac{1}{n+1}right) - n lnleft(1 + frac{1}{n}right) )I need to show that this difference is positive.Maybe I can approximate the logarithm terms using the Taylor series expansion. The natural logarithm can be approximated as:( ln(1 + x) approx x - frac{x^2}{2} + frac{x^3}{3} - dots ) for small ( x ).Since ( frac{1}{n} ) and ( frac{1}{n+1} ) are small for large ( n ), this approximation might be useful.Let me apply this approximation to both logarithm terms:( lnleft(1 + frac{1}{n}right) approx frac{1}{n} - frac{1}{2n^2} + frac{1}{3n^3} - dots )Similarly,( lnleft(1 + frac{1}{n+1}right) approx frac{1}{n+1} - frac{1}{2(n+1)^2} + frac{1}{3(n+1)^3} - dots )Now, plug these into the expression for ( ln(a_{n+1}) - ln(a_n) ):( (n+1)left(frac{1}{n+1} - frac{1}{2(n+1)^2} + frac{1}{3(n+1)^3} - dotsright) - nleft(frac{1}{n} - frac{1}{2n^2} + frac{1}{3n^3} - dotsright) )Simplify each term:First term:( (n+1)left(frac{1}{n+1}right) = 1 )( (n+1)left(-frac{1}{2(n+1)^2}right) = -frac{1}{2(n+1)} )( (n+1)left(frac{1}{3(n+1)^3}right) = frac{1}{3(n+1)^2} )And so on.Second term:( nleft(frac{1}{n}right) = 1 )( nleft(-frac{1}{2n^2}right) = -frac{1}{2n} )( nleft(frac{1}{3n^3}right) = frac{1}{3n^2} )And so on.Now, subtract the second expression from the first:( left[1 - frac{1}{2(n+1)} + frac{1}{3(n+1)^2} - dotsright] - left[1 - frac{1}{2n} + frac{1}{3n^2} - dotsright] )Simplify:( 1 - frac{1}{2(n+1)} + frac{1}{3(n+1)^2} - dots - 1 + frac{1}{2n} - frac{1}{3n^2} + dots )The 1's cancel out:( -frac{1}{2(n+1)} + frac{1}{3(n+1)^2} - dots + frac{1}{2n} - frac{1}{3n^2} + dots )Now, let's group the terms:( left(frac{1}{2n} - frac{1}{2(n+1)}right) + left(frac{1}{3(n+1)^2} - frac{1}{3n^2}right) + dots )Compute each group:First group:( frac{1}{2n} - frac{1}{2(n+1)} = frac{1}{2}left(frac{1}{n} - frac{1}{n+1}right) = frac{1}{2}left(frac{1}{n(n+1)}right) = frac{1}{2n(n+1)} )Second group:( frac{1}{3(n+1)^2} - frac{1}{3n^2} = frac{1}{3}left(frac{1}{(n+1)^2} - frac{1}{n^2}right) = frac{1}{3}left(frac{n^2 - (n+1)^2}{n^2(n+1)^2}right) = frac{1}{3}left(frac{-2n -1}{n^2(n+1)^2}right) = -frac{2n +1}{3n^2(n+1)^2} )So, putting it all together:( frac{1}{2n(n+1)} - frac{2n +1}{3n^2(n+1)^2} + dots )Now, let's see if this is positive. The first term is positive, and the second term is negative. I need to check if the positive term outweighs the negative term.Compute the difference:( frac{1}{2n(n+1)} - frac{2n +1}{3n^2(n+1)^2} )Let me factor out ( frac{1}{n(n+1)} ):( frac{1}{n(n+1)}left(frac{1}{2} - frac{2n +1}{3n(n+1)}right) )Simplify the expression inside the parentheses:( frac{1}{2} - frac{2n +1}{3n(n+1)} )Let me compute this:First, find a common denominator, which is 6n(n+1):( frac{3n(n+1)}{6n(n+1)} - frac{2(2n +1)}{6n(n+1)} )Simplify:( frac{3n(n+1) - 4n - 2}{6n(n+1)} )Expand ( 3n(n+1) ):( 3n^2 + 3n - 4n - 2 = 3n^2 - n - 2 )So, the expression becomes:( frac{3n^2 - n - 2}{6n(n+1)} )Now, factor the numerator:( 3n^2 - n - 2 )Looking for two numbers that multiply to ( 3 times (-2) = -6 ) and add to -1. Those numbers are -3 and 2.So, split the middle term:( 3n^2 - 3n + 2n - 2 )Factor by grouping:( 3n(n - 1) + 2(n - 1) = (3n + 2)(n - 1) )Wait, that doesn't seem right. Let me check:( 3n^2 - 3n + 2n - 2 = 3n(n - 1) + 2(n - 1) = (3n + 2)(n - 1) )Yes, that's correct.So, the numerator factors as ( (3n + 2)(n - 1) ).Therefore, the expression becomes:( frac{(3n + 2)(n - 1)}{6n(n+1)} )Now, since ( n ) is a positive integer greater than or equal to 1, ( n - 1 ) is non-negative for ( n geq 1 ), and ( 3n + 2 ) is always positive.Thus, the entire expression inside the parentheses is non-negative for ( n geq 1 ).Therefore, the difference ( ln(a_{n+1}) - ln(a_n) ) is positive, meaning ( a_{n+1} > a_n ), so the sequence ( a_n ) is increasing.Okay, that took a while, but I think I managed to show that ( a_n ) is increasing.Now, moving on to the sequence ( b_n = left(1 + frac{1}{n}right)^{n+1} ). I need to show that this sequence is decreasing.Similarly, I can take the ratio ( frac{b_{n+1}}{b_n} ) and show that it's less than 1.So,( frac{b_{n+1}}{b_n} = frac{left(1 + frac{1}{n+1}right)^{n+2}}{left(1 + frac{1}{n}right)^{n+1}} )Again, taking logarithms might help.Define ( ln(b_n) = (n+1) lnleft(1 + frac{1}{n}right) )Then,( ln(b_{n+1}) - ln(b_n) = (n+2) lnleft(1 + frac{1}{n+1}right) - (n+1) lnleft(1 + frac{1}{n}right) )I need to show that this difference is negative.Using the same approach as before, I'll approximate the logarithms using the Taylor series.( lnleft(1 + frac{1}{n}right) approx frac{1}{n} - frac{1}{2n^2} + frac{1}{3n^3} - dots )( lnleft(1 + frac{1}{n+1}right) approx frac{1}{n+1} - frac{1}{2(n+1)^2} + frac{1}{3(n+1)^3} - dots )Plugging these into the expression:( (n+2)left(frac{1}{n+1} - frac{1}{2(n+1)^2} + frac{1}{3(n+1)^3} - dotsright) - (n+1)left(frac{1}{n} - frac{1}{2n^2} + frac{1}{3n^3} - dotsright) )Simplify each term:First term:( (n+2)left(frac{1}{n+1}right) = frac{n+2}{n+1} = 1 + frac{1}{n+1} )( (n+2)left(-frac{1}{2(n+1)^2}right) = -frac{n+2}{2(n+1)^2} )( (n+2)left(frac{1}{3(n+1)^3}right) = frac{n+2}{3(n+1)^3} )And so on.Second term:( (n+1)left(frac{1}{n}right) = 1 + frac{1}{n} )( (n+1)left(-frac{1}{2n^2}right) = -frac{n+1}{2n^2} )( (n+1)left(frac{1}{3n^3}right) = frac{n+1}{3n^3} )And so on.Now, subtract the second expression from the first:( left[1 + frac{1}{n+1} - frac{n+2}{2(n+1)^2} + frac{n+2}{3(n+1)^3} - dotsright] - left[1 + frac{1}{n} - frac{n+1}{2n^2} + frac{n+1}{3n^3} - dotsright] )Simplify:( 1 + frac{1}{n+1} - frac{n+2}{2(n+1)^2} + frac{n+2}{3(n+1)^3} - dots - 1 - frac{1}{n} + frac{n+1}{2n^2} - frac{n+1}{3n^3} + dots )The 1's cancel out:( frac{1}{n+1} - frac{n+2}{2(n+1)^2} + frac{n+2}{3(n+1)^3} - dots - frac{1}{n} + frac{n+1}{2n^2} - frac{n+1}{3n^3} + dots )Now, group similar terms:( left(frac{1}{n+1} - frac{1}{n}right) + left(-frac{n+2}{2(n+1)^2} + frac{n+1}{2n^2}right) + left(frac{n+2}{3(n+1)^3} - frac{n+1}{3n^3}right) + dots )Compute each group:First group:( frac{1}{n+1} - frac{1}{n} = -frac{1}{n(n+1)} )Second group:( -frac{n+2}{2(n+1)^2} + frac{n+1}{2n^2} = frac{1}{2}left(-frac{n+2}{(n+1)^2} + frac{n+1}{n^2}right) )Let me compute this:( -frac{n+2}{(n+1)^2} + frac{n+1}{n^2} = -frac{n+2}{(n+1)^2} + frac{n+1}{n^2} )To combine these, find a common denominator, which is ( n^2(n+1)^2 ):( -frac{(n+2)n^2}{n^2(n+1)^2} + frac{(n+1)^3}{n^2(n+1)^2} = frac{-(n+2)n^2 + (n+1)^3}{n^2(n+1)^2} )Expand ( (n+1)^3 ):( (n+1)^3 = n^3 + 3n^2 + 3n + 1 )Expand ( -(n+2)n^2 ):( -(n+2)n^2 = -n^3 - 2n^2 )Combine the two:( (-n^3 - 2n^2) + (n^3 + 3n^2 + 3n + 1) = ( -n^3 + n^3 ) + ( -2n^2 + 3n^2 ) + 3n + 1 = n^2 + 3n + 1 )So, the second group becomes:( frac{n^2 + 3n + 1}{2n^2(n+1)^2} )Third group:( frac{n+2}{3(n+1)^3} - frac{n+1}{3n^3} = frac{1}{3}left(frac{n+2}{(n+1)^3} - frac{n+1}{n^3}right) )Again, find a common denominator ( n^3(n+1)^3 ):( frac{(n+2)n^3}{n^3(n+1)^3} - frac{(n+1)^4}{n^3(n+1)^3} = frac{(n+2)n^3 - (n+1)^4}{n^3(n+1)^3} )Expand ( (n+1)^4 ):( (n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n + 1 )Expand ( (n+2)n^3 ):( (n+2)n^3 = n^4 + 2n^3 )Subtract:( (n^4 + 2n^3) - (n^4 + 4n^3 + 6n^2 + 4n + 1) = -2n^3 - 6n^2 - 4n - 1 )So, the third group becomes:( frac{-2n^3 - 6n^2 - 4n - 1}{3n^3(n+1)^3} )Putting it all together:( -frac{1}{n(n+1)} + frac{n^2 + 3n + 1}{2n^2(n+1)^2} - frac{2n^3 + 6n^2 + 4n + 1}{3n^3(n+1)^3} + dots )This is getting quite complicated. Maybe instead of expanding all the terms, I can consider the behavior as ( n ) increases.Alternatively, perhaps I can use the fact that ( b_n = left(1 + frac{1}{n}right)^{n+1} ) and relate it to ( a_n ).Since ( a_n = left(1 + frac{1}{n}right)^n ), then ( b_n = a_n left(1 + frac{1}{n}right) ).So, ( b_n = a_n cdot left(1 + frac{1}{n}right) ).Given that ( a_n ) is increasing, and ( left(1 + frac{1}{n}right) ) is decreasing, the product ( b_n ) might be decreasing.But I need to formalize this.Alternatively, consider the ratio ( frac{b_{n+1}}{b_n} ):( frac{b_{n+1}}{b_n} = frac{left(1 + frac{1}{n+1}right)^{n+2}}{left(1 + frac{1}{n}right)^{n+1}} )Let me write this as:( left(frac{n+2}{n+1}right)^{n+2} cdot left(frac{n}{n+1}right)^{n+1} )Wait, that might not be helpful.Alternatively, take the natural logarithm:( lnleft(frac{b_{n+1}}{b_n}right) = (n+2)lnleft(1 + frac{1}{n+1}right) - (n+1)lnleft(1 + frac{1}{n}right) )I need to show that this is negative.Let me denote ( f(n) = (n+1)lnleft(1 + frac{1}{n}right) ). Then, ( ln(b_n) = f(n) ).So, ( lnleft(frac{b_{n+1}}{b_n}right) = f(n+1) - f(n) )If I can show that ( f(n) ) is decreasing, then ( f(n+1) - f(n) < 0 ), which would imply ( b_{n+1} < b_n ).So, let's analyze ( f(n) = (n+1)lnleft(1 + frac{1}{n}right) ).Compute the difference ( f(n+1) - f(n) ):( (n+2)lnleft(1 + frac{1}{n+1}right) - (n+1)lnleft(1 + frac{1}{n}right) )I need to show this is negative.Again, using the Taylor series approximation for ( ln(1 + x) ):( lnleft(1 + frac{1}{n}right) approx frac{1}{n} - frac{1}{2n^2} + frac{1}{3n^3} - dots )Similarly,( lnleft(1 + frac{1}{n+1}right) approx frac{1}{n+1} - frac{1}{2(n+1)^2} + frac{1}{3(n+1)^3} - dots )Plugging these into ( f(n+1) - f(n) ):( (n+2)left(frac{1}{n+1} - frac{1}{2(n+1)^2} + frac{1}{3(n+1)^3} - dotsright) - (n+1)left(frac{1}{n} - frac{1}{2n^2} + frac{1}{3n^3} - dotsright) )Simplify each term:First term:( (n+2)left(frac{1}{n+1}right) = 1 + frac{1}{n+1} )( (n+2)left(-frac{1}{2(n+1)^2}right) = -frac{n+2}{2(n+1)^2} )( (n+2)left(frac{1}{3(n+1)^3}right) = frac{n+2}{3(n+1)^3} )Second term:( (n+1)left(frac{1}{n}right) = 1 + frac{1}{n} )( (n+1)left(-frac{1}{2n^2}right) = -frac{n+1}{2n^2} )( (n+1)left(frac{1}{3n^3}right) = frac{n+1}{3n^3} )Now, subtract the second expression from the first:( left[1 + frac{1}{n+1} - frac{n+2}{2(n+1)^2} + frac{n+2}{3(n+1)^3} - dotsright] - left[1 + frac{1}{n} - frac{n+1}{2n^2} + frac{n+1}{3n^3} - dotsright] )Simplify:( 1 + frac{1}{n+1} - frac{n+2}{2(n+1)^2} + frac{n+2}{3(n+1)^3} - dots - 1 - frac{1}{n} + frac{n+1}{2n^2} - frac{n+1}{3n^3} + dots )The 1's cancel out:( frac{1}{n+1} - frac{n+2}{2(n+1)^2} + frac{n+2}{3(n+1)^3} - dots - frac{1}{n} + frac{n+1}{2n^2} - frac{n+1}{3n^3} + dots )Group similar terms:( left(frac{1}{n+1} - frac{1}{n}right) + left(-frac{n+2}{2(n+1)^2} + frac{n+1}{2n^2}right) + left(frac{n+2}{3(n+1)^3} - frac{n+1}{3n^3}right) + dots )Compute each group:First group:( frac{1}{n+1} - frac{1}{n} = -frac{1}{n(n+1)} )Second group:( -frac{n+2}{2(n+1)^2} + frac{n+1}{2n^2} = frac{1}{2}left(-frac{n+2}{(n+1)^2} + frac{n+1}{n^2}right) )Let me compute this:( -frac{n+2}{(n+1)^2} + frac{n+1}{n^2} = -frac{n+2}{(n+1)^2} + frac{n+1}{n^2} )Find a common denominator ( n^2(n+1)^2 ):( -frac{(n+2)n^2}{n^2(n+1)^2} + frac{(n+1)^3}{n^2(n+1)^2} = frac{-(n+2)n^2 + (n+1)^3}{n^2(n+1)^2} )Expand ( (n+1)^3 ):( (n+1)^3 = n^3 + 3n^2 + 3n + 1 )Expand ( -(n+2)n^2 ):( -(n+2)n^2 = -n^3 - 2n^2 )Combine the two:( (-n^3 - 2n^2) + (n^3 + 3n^2 + 3n + 1) = ( -n^3 + n^3 ) + ( -2n^2 + 3n^2 ) + 3n + 1 = n^2 + 3n + 1 )So, the second group becomes:( frac{n^2 + 3n + 1}{2n^2(n+1)^2} )Third group:( frac{n+2}{3(n+1)^3} - frac{n+1}{3n^3} = frac{1}{3}left(frac{n+2}{(n+1)^3} - frac{n+1}{n^3}right) )Again, find a common denominator ( n^3(n+1)^3 ):( frac{(n+2)n^3}{n^3(n+1)^3} - frac{(n+1)^4}{n^3(n+1)^3} = frac{(n+2)n^3 - (n+1)^4}{n^3(n+1)^3} )Expand ( (n+1)^4 ):( (n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n + 1 )Expand ( (n+2)n^3 ):( (n+2)n^3 = n^4 + 2n^3 )Subtract:( (n^4 + 2n^3) - (n^4 + 4n^3 + 6n^2 + 4n + 1) = -2n^3 - 6n^2 - 4n - 1 )So, the third group becomes:( frac{-2n^3 - 6n^2 - 4n - 1}{3n^3(n+1)^3} )Putting it all together:( -frac{1}{n(n+1)} + frac{n^2 + 3n + 1}{2n^2(n+1)^2} - frac{2n^3 + 6n^2 + 4n + 1}{3n^3(n+1)^3} + dots )This is quite involved. Maybe instead of expanding all the terms, I can consider the dominant terms.The first term is negative, the second term is positive, and the third term is negative. I need to see if the overall expression is negative.Alternatively, perhaps I can consider the function ( f(n) = (n+1)lnleft(1 + frac{1}{n}right) ) and analyze its behavior as ( n ) increases.Compute the derivative of ( f(n) ) with respect to ( n ):( f(n) = (n+1)lnleft(1 + frac{1}{n}right) )Let me write ( x = n ), so ( f(x) = (x+1)lnleft(1 + frac{1}{x}right) )Compute ( f'(x) ):Using the product rule:( f'(x) = lnleft(1 + frac{1}{x}right) + (x+1) cdot frac{d}{dx}lnleft(1 + frac{1}{x}right) )Compute the derivative inside:( frac{d}{dx}lnleft(1 + frac{1}{x}right) = frac{1}{1 + frac{1}{x}} cdot left(-frac{1}{x^2}right) = frac{-1}{x^2 + x} )So,( f'(x) = lnleft(1 + frac{1}{x}right) - frac{x+1}{x^2 + x} )Simplify ( frac{x+1}{x^2 + x} = frac{x+1}{x(x+1)} = frac{1}{x} )Thus,( f'(x) = lnleft(1 + frac{1}{x}right) - frac{1}{x} )Now, I need to determine the sign of ( f'(x) ).Recall that for ( x > 0 ), ( ln(1 + y) < y ) for ( y > 0 ). Here, ( y = frac{1}{x} ), so ( lnleft(1 + frac{1}{x}right) < frac{1}{x} )Therefore, ( f'(x) = lnleft(1 + frac{1}{x}right) - frac{1}{x} < 0 )This means that ( f(x) ) is decreasing for ( x > 0 ). Therefore, ( f(n+1) < f(n) ), which implies ( ln(b_{n+1}) - ln(b_n) < 0 ), so ( b_{n+1} < b_n ). Hence, the sequence ( b_n ) is decreasing.Great, that seems to work.Now, the problem also states that any element of ( b_n ) is greater than all elements of ( a_n ). That is, for any ( m ) and ( k ), ( b_m > a_k ).Given that ( a_n ) is increasing and ( b_n ) is decreasing, and both converge to ( e ), it makes sense that every ( b_n ) is greater than every ( a_n ).But to formalize this, perhaps consider that for any ( n ), ( b_n = a_n cdot left(1 + frac{1}{n}right) ). Since ( a_n ) is increasing and ( left(1 + frac{1}{n}right) > 1 ), then ( b_n > a_n ).Moreover, since ( a_n ) is increasing, ( a_k leq a_n ) for ( k leq n ), and since ( b_n ) is decreasing, ( b_n geq b_m ) for ( m geq n ). Therefore, for any ( m ) and ( k ), choosing ( n ) such that ( n geq k ) and ( n geq m ), we have ( a_k leq a_n < b_n leq b_m ). Hence, ( a_k < b_m ).Finally, the problem asks to show that the function ( f(x) = left(1 + frac{1}{x}right)^x ) increases as ( x ) increases.We have already shown that for integer ( n ), ( a_n ) is increasing. To extend this to the function ( f(x) ), we can consider the derivative of ( f(x) ) with respect to ( x ).Let me compute ( f(x) = left(1 + frac{1}{x}right)^x )Take the natural logarithm:( ln(f(x)) = x lnleft(1 + frac{1}{x}right) )Differentiate both sides with respect to ( x ):( frac{f'(x)}{f(x)} = lnleft(1 + frac{1}{x}right) + x cdot frac{d}{dx}lnleft(1 + frac{1}{x}right) )We already computed this derivative earlier:( frac{d}{dx}lnleft(1 + frac{1}{x}right) = -frac{1}{x^2 + x} )So,( frac{f'(x)}{f(x)} = lnleft(1 + frac{1}{x}right) - frac{x}{x^2 + x} = lnleft(1 + frac{1}{x}right) - frac{1}{x + 1} )We need to determine the sign of ( f'(x) ). Since ( f(x) > 0 ), the sign of ( f'(x) ) is the same as the sign of ( lnleft(1 + frac{1}{x}right) - frac{1}{x + 1} ).Let me denote ( g(x) = lnleft(1 + frac{1}{x}right) - frac{1}{x + 1} )I need to show that ( g(x) > 0 ) for ( x > 0 ).Consider the function ( h(y) = ln(1 + y) - frac{y}{1 + y} ) where ( y = frac{1}{x} ), so ( y > 0 ).Compute ( h(y) = ln(1 + y) - frac{y}{1 + y} )Compute the derivative ( h'(y) ):( h'(y) = frac{1}{1 + y} - frac{(1)(1 + y) - y(1)}{(1 + y)^2} = frac{1}{1 + y} - frac{1}{(1 + y)^2} = frac{(1 + y) - 1}{(1 + y)^2} = frac{y}{(1 + y)^2} > 0 ) for ( y > 0 )Since ( h'(y) > 0 ), ( h(y) ) is increasing for ( y > 0 ).Compute ( h(0) = ln(1 + 0) - frac{0}{1 + 0} = 0 )Since ( h(y) ) is increasing and ( h(0) = 0 ), then for ( y > 0 ), ( h(y) > 0 ). Therefore, ( g(x) = hleft(frac{1}{x}right) > 0 ) for ( x > 0 ).Thus, ( f'(x) > 0 ), meaning ( f(x) ) is increasing for ( x > 0 ).So, putting it all together, we've shown that:1. ( lg left(1+frac{1}{n}right)^{n} ) increases as ( n ) increases because ( a_n ) is increasing.2. ( lg left(1+frac{1}{n}right)^{n+1} ) decreases as ( n ) increases because ( b_n ) is decreasing.3. The sequence ( b_n ) is decreasing and every term of ( b_n ) is greater than every term of ( a_n ).4. The function ( left(1 + frac{1}{x}right)^x ) is increasing for all ( x > 0 ).I think I've covered all the parts of the problem. It was a bit lengthy, especially with all the approximations and derivative computations, but breaking it down step by step helped me understand each part better.

💡First, I need to determine how long it took Cheenu to run one mile when he was a young man. He ran 20 miles in 4 hours. To find the time per mile, I'll convert the total time into minutes and then divide by the number of miles.Next, I'll calculate how long it takes him to walk one mile now that he's older. He walks 12 miles in 5 hours. Again, I'll convert the total time into minutes and divide by the number of miles to find the time per mile.Finally, I'll find the difference between the current time per mile and the time per mile when he was younger. This difference will tell me how many more minutes it takes him to walk a mile now compared to when he was a young man.

💡Let ( n ) be the number of members in the choir. According to the problem:1. When arranged in 10 rows, there are 4 positions unoccupied. This implies: [ n equiv 6 pmod{10} ] (since 10 - 4 = 6 to fill up the 10th row).2. When arranged in 11 rows, there are 5 positions unoccupied. This implies: [ n equiv 6 pmod{11} ] (since 11 - 5 = 6 to fill up the 11th row).To find a number ( n ) that satisfies both conditions, we calculate the least common multiple (LCM) of 10 and 11: [ text{LCM of 10 and 11} = 110 ]Thus, ( n ) must be 6 more than a multiple of 110: [ n = 110k + 6 ]where ( k ) is an integer. We need ( n ) to be between 200 and 300: [ 200 leq 110k + 6 leq 300 ] [ 194 leq 110k leq 294 ] [ 1.76364 leq k leq 2.67273 ]The only integer value of ( k ) within this range is ( k = 2 ). Substituting back, we find: [ n = 110 times 2 + 6 = 226 ]Therefore, the number of members in the choir is (boxed{226}).

💡First, I need to identify the diagonals of the rhombus. The vertical diagonal connects the points (0, 3.5) and (0, -3.5), which is 7 units long.Next, the horizontal diagonal connects the points (-8, 0) and (x, 0). The length of this diagonal is x + 8 units.Using the area formula for a rhombus, which is (diagonal1 * diagonal2) / 2, and knowing the area is 56, I can set up the equation:(7 * (x + 8)) / 2 = 56Solving for x:7 * (x + 8) = 112x + 8 = 16x = 8Therefore, the missing x-coordinate is 8.

💡Alright, so I have this functional equation to solve: ( f(x^2 - y^2) = (x - y)(f(x) + f(y)) ) for all real numbers ( x ) and ( y ). I need to find all functions ( f: mathbb{R} rightarrow mathbb{R} ) that satisfy this condition. Hmm, functional equations can be tricky, but let me try to break it down step by step.First, I should check if there are any obvious solutions. Sometimes, simple functions like linear functions or the zero function might work. Let me try ( f(x) = 0 ). Plugging that into the equation:Left side: ( f(x^2 - y^2) = 0 ).Right side: ( (x - y)(0 + 0) = 0 ).Okay, that works. So ( f(x) = 0 ) is a solution. What about a linear function, say ( f(x) = kx ) where ( k ) is a constant? Let's test that.Left side: ( f(x^2 - y^2) = k(x^2 - y^2) ).Right side: ( (x - y)(kx + ky) = k(x - y)(x + y) = k(x^2 - y^2) ).That also works! So any linear function ( f(x) = kx ) satisfies the equation. Interesting. So maybe all solutions are linear functions of this form?But I need to be sure there aren't any other solutions. Let me try to explore further.Let me consider specific values for ( x ) and ( y ) to see if I can get more information about ( f ). A common strategy is to set one of the variables to zero or make them equal.First, let's set ( x = y = 0 ):Left side: ( f(0^2 - 0^2) = f(0) ).Right side: ( (0 - 0)(f(0) + f(0)) = 0 ).So, ( f(0) = 0 ). That's a useful piece of information.Next, let's set ( y = 0 ) and see what happens:Left side: ( f(x^2 - 0^2) = f(x^2) ).Right side: ( (x - 0)(f(x) + f(0)) = x(f(x) + 0) = x f(x) ).So, ( f(x^2) = x f(x) ). That's an interesting relation. Let me note that down: ( f(x^2) = x f(x) ).Similarly, let's set ( x = 0 ) and see:Left side: ( f(0^2 - y^2) = f(-y^2) ).Right side: ( (0 - y)(f(0) + f(y)) = (-y)(0 + f(y)) = -y f(y) ).So, ( f(-y^2) = -y f(y) ). Hmm, that's another relation. Let me see if I can relate this to the previous one.From the first relation, ( f(x^2) = x f(x) ). If I replace ( x ) with ( -x ), I get ( f((-x)^2) = (-x) f(-x) ). But ( (-x)^2 = x^2 ), so ( f(x^2) = -x f(-x) ).But from the first relation, ( f(x^2) = x f(x) ). Therefore, ( x f(x) = -x f(-x) ). Assuming ( x neq 0 ), we can divide both sides by ( x ):( f(x) = -f(-x) ).So, ( f ) is an odd function. That's a useful property. It means that ( f(-x) = -f(x) ) for all ( x ).Okay, so far, I have:1. ( f(0) = 0 ).2. ( f(x^2) = x f(x) ).3. ( f(-y^2) = -y f(y) ).4. ( f ) is odd: ( f(-x) = -f(x) ).Let me see if I can use these properties to find more about ( f ).Going back to the original equation: ( f(x^2 - y^2) = (x - y)(f(x) + f(y)) ).I wonder if I can express ( f(x^2 - y^2) ) in terms of ( f(x^2) ) and ( f(y^2) ). From the relation ( f(x^2) = x f(x) ), I can write:( f(x^2 - y^2) = f(x^2) + f(-y^2) ) if ( f ) is additive. Wait, but I don't know if ( f ) is additive. Hmm, maybe that's not the right approach.Alternatively, since ( f ) is odd, ( f(-y^2) = -f(y^2) ). So, ( f(x^2 - y^2) = f(x^2) + f(-y^2) = f(x^2) - f(y^2) ).But from the original equation, ( f(x^2 - y^2) = (x - y)(f(x) + f(y)) ).So, combining these two expressions:( f(x^2) - f(y^2) = (x - y)(f(x) + f(y)) ).But from the relation ( f(x^2) = x f(x) ), we can substitute:( x f(x) - y f(y) = (x - y)(f(x) + f(y)) ).Let me write that down:( x f(x) - y f(y) = (x - y)(f(x) + f(y)) ).Let me expand the right side:( (x - y)(f(x) + f(y)) = x f(x) + x f(y) - y f(x) - y f(y) ).So, the equation becomes:( x f(x) - y f(y) = x f(x) + x f(y) - y f(x) - y f(y) ).Let me subtract ( x f(x) - y f(y) ) from both sides to see what's left:( 0 = x f(y) - y f(x) ).So, ( x f(y) = y f(x) ).Hmm, that's an interesting relation. Let me write that:( x f(y) = y f(x) ) for all ( x, y in mathbb{R} ).This seems like a key equation. Let me see what this implies.If I fix ( y = 1 ), then:( x f(1) = 1 f(x) ).So, ( f(x) = x f(1) ).That suggests that ( f ) is a linear function proportional to ( x ), with the constant of proportionality ( f(1) ). Let me denote ( k = f(1) ), so ( f(x) = k x ).Wait, that's exactly the solution I tested earlier! So, this suggests that the only solutions are linear functions of the form ( f(x) = k x ).But let me make sure there are no other solutions. Suppose ( f ) is not linear, but satisfies ( x f(y) = y f(x) ) for all ( x, y ). Is this possible?Let me think. Suppose ( f ) is not linear, but satisfies ( x f(y) = y f(x) ). Let me rearrange this equation:( frac{f(x)}{x} = frac{f(y)}{y} ) for all ( x, y neq 0 ).This suggests that ( frac{f(x)}{x} ) is a constant for all ( x neq 0 ). Let me denote this constant by ( k ), so ( f(x) = k x ) for all ( x neq 0 ). Since ( f(0) = 0 ), this holds for all ( x ).Therefore, the only solutions are linear functions ( f(x) = k x ).Let me verify this again. Suppose ( f(x) = k x ). Then,Left side: ( f(x^2 - y^2) = k(x^2 - y^2) ).Right side: ( (x - y)(f(x) + f(y)) = (x - y)(k x + k y) = k(x - y)(x + y) = k(x^2 - y^2) ).Yes, both sides are equal. So, indeed, ( f(x) = k x ) satisfies the equation.I think I've covered all possible cases. I started by checking obvious solutions, then used specific values to derive properties of ( f ), and finally arrived at the conclusion that ( f ) must be linear. I don't see any other possibilities, so I think the only solutions are linear functions of the form ( f(x) = k x ).

💡Alright, so I have this combination lock problem with 6 rotating discs, each labeled from 0 to 9. The lock starts at 000000, and I know that's not the correct combination. I need to figure out the least number of moves required to guarantee that I find the correct combination. A move is turning one disc one digit in either direction.First, let me understand the problem better. Each disc has 10 possible digits, and since there are 6 discs, the total number of possible combinations is 10^6, which is 1,000,000. That's a lot! But since we know that 000000 isn't the correct one, we have 999,999 possible combinations left.Now, the question is asking for the least number of moves necessary to ensure that we've tried every possible combination. Each move allows me to change one disc by one digit. So, to go from 000000 to 000001, that's one move. To go from 000001 to 000002, that's another move, and so on.But wait, if I have to try every single combination, how many moves would that take? If I start at 000000 and move each disc one by one, incrementing each digit, it would take a lot of moves. But maybe there's a smarter way to do this, like a sequence that covers all combinations with minimal moves.I remember something about Hamiltonian paths in graphs, where each node represents a combination, and edges represent a single move. If I can traverse all nodes with a path that only moves one disc at a time, that would give me the minimal number of moves. But I'm not sure how to apply that here.Alternatively, maybe I can think of it as a covering problem. Each move covers one combination, and I need to cover all 999,999 combinations. So, in the worst case, I might need 999,999 moves, right? But that seems too straightforward.Wait, no. Because each move changes one disc, and each disc has 10 digits, maybe I can cycle through all the digits on one disc before moving on to the next. But that would still require a lot of moves.Let me think about a smaller case first. Suppose there are only 2 discs instead of 6. Each disc has 10 digits, so there are 100 combinations. Starting at 00, I need to find the correct combination. If I increment the first disc one by one, I'd go through 00, 10, 20, ..., 90, and then increment the second disc to 01, 11, 21, ..., 91, and so on. But that's not efficient because I'm only changing one disc at a time.Alternatively, if I alternate between the two discs, I can cover more combinations faster. But I'm not sure if that helps in minimizing the total number of moves.Wait, maybe the key is to realize that each move only changes one disc, so to get from one combination to another, you have to change each disc one by one. So, to cover all combinations, you have to make sure that every possible digit on every disc is tried in some sequence.But how can I ensure that? Maybe by cycling through each disc in a systematic way. For example, fix the first five discs and cycle through the sixth disc, then move to the fifth disc and cycle through it while keeping the others fixed, and so on.But that would still require a lot of moves. For each disc, you have to cycle through 10 digits, and there are 6 discs. So, 6 * 10 = 60 moves, but that's just to cycle through all digits on all discs once. But we have 1,000,000 combinations, so that's not enough.Wait, maybe I'm overcomplicating it. If I need to try every combination, and each move only changes one disc, then the minimal number of moves required is equal to the number of combinations minus one. Because starting from 000000, each move takes you to a new combination, so after n moves, you've tried n+1 combinations. Therefore, to try all 999,999 combinations, you need 999,998 moves.But that seems too simple. Is there a way to do it in fewer moves? Maybe by reusing some combinations or overlapping moves? But I don't think so because each move only changes one disc, so you can't cover multiple combinations in a single move.Wait, but what if the correct combination is found before trying all of them? The question says "to ensure that we have found the correct combination," so we need to consider the worst-case scenario, where the correct combination is the last one we try. Therefore, we need to plan for the case where we have to try all possible combinations.So, if we start at 000000, and need to try 999,999 combinations, each requiring one move, then the total number of moves is 999,999. But since we start at 000000, which is not correct, the first move takes us to the first combination, and the last move takes us to the 999,999th combination. Therefore, the number of moves is 999,998.Wait, that makes sense. Because the number of moves is one less than the number of combinations you need to try. So, if you have to try 999,999 combinations, you need 999,998 moves.But let me double-check. If I have 2 combinations, I need 1 move to go from the first to the second. If I have 3 combinations, I need 2 moves. So, yes, the number of moves is always one less than the number of combinations you need to try.Therefore, for part (a), the least number of moves necessary is 999,998.Now, moving on to part (b). It says that none of the combinations 000000, 111111, 222222, ..., 999999 is correct. So, these 10 combinations are invalid. That means we have 1,000,000 - 10 = 999,990 valid combinations.But wait, does that change the number of moves? Because we still need to try all the valid combinations, but we can skip the 10 invalid ones. So, instead of trying 999,999 combinations, we only need to try 999,990.But hold on, the way the lock works is that it opens instantly when the correct combination is entered. So, if we follow a sequence of moves that covers all combinations, and we know that certain combinations are invalid, can we adjust our sequence to skip those invalid ones?But the problem is, we don't know which combinations are invalid except for those 10 all-same-digit combinations. So, in terms of planning, we still have to consider that the correct combination could be any of the remaining 999,990.But does that affect the minimal number of moves? Because in the worst case, the correct combination could still be the last one we try, so we still need to plan for trying all 999,990 combinations. But how does that translate to the number of moves?Wait, similar to part (a), the number of moves is one less than the number of combinations we need to try. So, if we have 999,990 combinations, we need 999,989 moves.But that contradicts my earlier conclusion. Wait, no, because in part (a), we had 999,999 combinations to try, requiring 999,998 moves. Here, we have 999,990 combinations, so 999,989 moves.But the answer might still be the same because the difference is negligible compared to the total number. Or maybe not.Wait, let me think again. The initial position is 000000, which is invalid. So, the first move takes us to a valid combination. Then, each subsequent move takes us to another combination. So, the number of moves is equal to the number of combinations we need to try minus one.Therefore, if we have 999,990 valid combinations, the number of moves is 999,989.But the problem is, we don't know which combinations are invalid except for those 10. So, in our sequence of moves, we have to make sure that we don't land on those 10 invalid combinations. But how?Because if we follow a sequence that covers all combinations, we might accidentally land on an invalid one, which we know is not correct, but the lock won't open, so we have to continue.But does that affect the total number of moves? Because even if we land on an invalid combination, we still have to continue trying others. So, in the worst case, we might have to try all 999,999 combinations, but we know that 10 of them are invalid, so we can potentially skip them.But how can we skip them? Because we don't know in advance which combinations are invalid, except for the 10 all-same-digit ones. So, in our sequence, we can plan to skip those 10 combinations.Therefore, instead of trying 999,999 combinations, we only need to try 999,999 - 10 = 999,989 combinations.Therefore, the number of moves is 999,988.Wait, but that seems inconsistent with my earlier reasoning. Let me clarify.If we have to try 999,990 valid combinations, starting from 000000, which is invalid, then the number of moves is 999,990 - 1 = 999,989.But if we can skip the 10 invalid combinations, then the number of moves is 999,990 - 1 = 999,989.Wait, but in part (a), we had 999,999 combinations to try, requiring 999,998 moves. Here, we have 999,990 combinations to try, requiring 999,989 moves.But the problem is, we don't know the correct combination, so we have to try all possible ones except the 10 invalid ones. Therefore, the number of moves is 999,990 - 1 = 999,989.But the answer might still be the same as part (a) because the difference is minimal, but I think it's actually less.Wait, no. Because in part (a), we had to try 999,999 combinations, requiring 999,998 moves. Here, we have to try 999,990 combinations, requiring 999,989 moves.But the question is, can we actually skip those 10 combinations? Because if we follow a sequence that covers all combinations, we might have to pass through those 10 invalid ones, which would waste moves.But if we can plan a sequence that skips those 10 combinations, then we can reduce the number of moves.But how?Maybe by using some kind of parity or pattern to avoid those combinations.Wait, the invalid combinations are all the ones where all digits are the same, like 000000, 111111, etc. So, maybe we can design a sequence that never has all digits the same.But how?Alternatively, maybe we can use the fact that the invalid combinations are all the same digit, so we can ensure that in our sequence, at least one digit is different from the others.But I'm not sure how to translate that into a sequence of moves.Alternatively, maybe we can think of the problem in terms of graph theory again, where each node is a combination, and edges represent a single move. Then, the problem reduces to finding a Hamiltonian path that avoids the 10 invalid nodes.But finding such a path is non-trivial, and I'm not sure if it's possible.Alternatively, maybe the number of moves remains the same because the difference is negligible, but I think it's actually less.Wait, let me think differently. If we know that 10 specific combinations are invalid, we can adjust our sequence to skip them. Therefore, instead of trying 999,999 combinations, we only need to try 999,999 - 10 = 999,989 combinations.Therefore, the number of moves is 999,989.But wait, in part (a), we started at 000000, which was invalid, and had to try 999,999 combinations, requiring 999,998 moves. Here, we start at 000000, which is invalid, and have to try 999,990 combinations, requiring 999,989 moves.Therefore, the answer for part (b) is 999,989.But I'm not entirely sure. Maybe I need to think about it differently.Alternatively, maybe the number of moves remains the same because the invalid combinations are spread out, and we can't really skip them without knowing where they are.But no, we do know where they are. We know that all combinations where all digits are the same are invalid. So, we can plan our sequence to avoid those.Therefore, the number of moves should be reduced by 10.But wait, in part (a), we had to try 999,999 combinations, requiring 999,998 moves. Here, we have to try 999,990 combinations, requiring 999,989 moves.So, the difference is 9 moves. That seems reasonable.Therefore, the answer for part (b) is 999,989.But I'm still a bit confused because the initial position is 000000, which is invalid, and we have to make moves to try other combinations. So, the number of moves is equal to the number of combinations we need to try minus one.Therefore, for part (a), 999,999 - 1 = 999,998 moves.For part (b), 999,990 - 1 = 999,989 moves.Yes, that makes sense.So, to summarize:a) The least number of moves necessary is 999,998.b) The least number of moves necessary is 999,989.