Find the series: ∑n=1∞((1+1n)n−e)

Question

Find the series:  ∑n=1∞((1+1n)n−e)

immablondevl · Accepted Answer

Take the negative of that series to make the terms positive and limit compare it with 1n  limn→∞e−(1+1n)n1n→00  ⇒limn→∞−(1+1n)n(log⁡(1+1n)−1n+1)−1n2  which we get from LHopital

gekraamdbk · Accepted Answer

We first observe, that  (1+1n)n−e=enln⁡(1+1n)−e  Elementary calculus arguments can show that the function xln⁡(1+1x) is monotone increasing on (0,∞) and that its range has 1 as a supremum and has 0 as an infimum.  We now invoke the Mean Value Theorem to conclude that for every n∈N there is a real ζn between 1 and nln⁡(1+1n) such that  e−(1+1n)n=eζn(1−nln⁡(1+1n))  From what we know about xln⁡(1+1x), we can conclude that  e0(1−nln⁡(1+1n))≤e−(1+1n)n≤e1(1−nln⁡(1+1n))  Now we use the inequalities  12(1+x)≤1−xln⁡(1+1x)≤1x+1  which is true for all positive x to conclude  12(1+n)≤e−(1+1n)n≤e1+n  which will carry the rest of the argument.  The real crux to this argument are the inequalities  12(1+x)≤1−xln⁡(1+1x)≤1x+1  The right-hand inequality is equivalent to the more familiar inequality  ln⁡(1+x)≤x  which I can derive if needed.  The left-hand inequality is equivalent to the slightly stronger inequality  ln⁡(1+x)≤x−x22(1+x)  which is however established in very much the same way as the weaker inequality.

Find the series: \sum_{n=1}^\infty((1+1/n)^n-e)

Answered question

Answer & Explanation

New Questions in Calculus 2