Consider the series

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series converges conditionally.

*Proof.* First, we use the Leibniz rule to show that the series converges. We have

so the terms are going to 0. Then, we need to show that the terms are monotonically decreasing. We will use the Arithmetic Mean-Geometric Mean inequality to do this (which we proved in this exercise, Section I.4.10, Exercise #20). This inequality states that for positive real numbers we have

(taking in the exercise) with equality only if . So, for the problem at hand we consider the positive real numbers

Then these are not all equal so we have strict inequality in the AM-GM inequality, giving us

Thus, the st term is greater than the th so the sequence whose terms are giving by is strictly increasing. Therefore, the sequence with terms

is strictly decreasing. Hence, by the Leibniz rule the series

converges.

This convergence is conditional since

Then, we use the limit comparison test to compare this series with the series . First, we make the substitution to obtain

Now, the limit of the first term in the product is

For the second term in the product we have

Therefore,

Hence, by the ratio test, the series diverges. So, the convergence of the series in the exercise is conditional

Hi, I am extremly interested how one may show, that the, on the first glance quite similar series of 1/n*(e-(1+1/n)^n) converges. I think the best criterium to use might be the one derived from cauchy-schwarzes inequality. using this what would be left to show is, that the series of (e-(1+1/n)^n)^2 convergrs, which i think it does. i know this questionen is only loosly connected to the problem at hand but i am really desperate to find out!

best regards

To show that (1+1/n)^n is monotonically increasing, you can also take the derivative of (1+1/x)^x.