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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.

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2 Answers

up vote 3 down vote accepted

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

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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? –  Mathlete Dec 1 '12 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 $n\gt1$) –  user127.0.0.1 Dec 1 '12 at 13:05
    
How is the limit 1? Should it not be greater than 1, in which case it would diverge, by the test. –  Mathlete Dec 1 '12 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? –  Mathlete Dec 1 '12 at 13:15
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$\frac{n}{n^2+1}=\frac{1}{n+1/n}\gt\frac{1}{n+1}$ –  user127.0.0.1 Dec 1 '12 at 13:17
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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.

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Yesiree!! ;-) +1 –  amWhy Apr 9 '13 at 0:32
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