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
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
Hence, by the ratio test, the series diverges. So, the convergence of the series in the exercise is conditional
In the last line, you said you used the ratio test but it’s actually the limit comparison test.
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!
To show that (1+1/n)^n is monotonically increasing, you can also take the derivative of (1+1/x)^x.