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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.

2 Answers 2

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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?2012-12-01
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    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 $n\gt1$)2012-12-01
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    How is the limit 1? Should it not be greater than 1, in which case it would diverge, by the test.2012-12-01
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    actually I see what you mean. How would I compare it to 1/n+1 though? What test would I need to use?2012-12-01
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    You have $\frac{n+1}{(n+1)^2+1}\frac{n^2+1}{n} = \frac{n^3+n^2+n+1}{n^3+2n^2+2n}\rightarrow 1$ so ratio test does not work here. I told you what to do in my answer!2012-12-01
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    $\frac{n}{n^2+1}=\frac{1}{n+1/n}\gt\frac{1}{n+1}$2012-12-01
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    I see, I just wanted to understand what exactly it was that I'm supposed to do.2012-12-01
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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.