# Determine whether this series is absolutely convergent, conditionally convergent or divergent?

The series $\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2 + 1}$; is it absolutely convergent, conditionally convergent or divergent?

This question is meant to be worth quite a few marks so although I thought I had the answer using the comparison test, I think I'm supposed to incorporate the alternating series test.

## 2 Answers

Your series is convergent by Leibniz-theorem but not absolutely convergent as you can see by comparison with $\sum \frac{1}{n+1}$

• I considered the modulus of the sequence $\frac{(-1)^n n}{n^2 + 1}$ which is $\frac{n}{n^2 + 1}$ and then found the limit of the modulus of $\frac{(a_(n+1))}{a_n}$ which I thought to be 0 - is that right? Commented Dec 1, 2012 at 12:59
• No, you should get 1 for that. Just write $\frac{n}{n^2+1} = \frac{1}{n+1/n}$ so it is clear that $a_n\rightarrow 0$. But you have to show, that $a_n$ is decreasing monotonously (for $n\gt1$) Commented Dec 1, 2012 at 13:05
• How is the limit 1? Should it not be greater than 1, in which case it would diverge, by the test. Commented Dec 1, 2012 at 13:08
• actually I see what you mean. How would I compare it to 1/n+1 though? What test would I need to use? Commented Dec 1, 2012 at 13:15
• $\frac{n}{n^2+1}=\frac{1}{n+1/n}\gt\frac{1}{n+1}$ Commented Dec 1, 2012 at 13:17

The way @Fant walked is practical, but maybe this approach also helps:

Use the integral test. As $f(x)=\frac{x}{x^2+1}$ is positive monotonic decreasing function on $x\geq 2$, so the integral test then $\sum_2^{\infty}f(n)$ converges or diverges if $\int_2^{\infty}f(x)dx$ converges or diverges. But the integral is clearly diverges, so we have here what @Fant noted again.