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