Consider the series

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

The given series converges absolutely.

*Proof.* First, we have

for all positive integers . (Since for .) Therefore,

Then, we write the Taylor expansion of ,

Thus,

But,

for since

Furthermore, . Hence,

therefore, by the comparison test

converges. So, finally, we have

converges. This implies

is absolutely convergent

Very nice. I think there is a typo in the solution here, and the convergence proved in exercise 10.9.3 seems more applicable in the solution here.

This is ridiculous! Did Apostol really want us to go crazy or is there a better solution? Thank you for your time RoRi

You could simply use the taylor expansion (or guess) that -A_n ~ 1/6x^2 and thus Σ-a_n converges. THen so does Σa_n