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I have to test if the series is absolute convergence and conditional convergence

$\sum_{n=1}^{\infty}$$\frac{(-1)^{n-1}n}{(n+1)^2}$

This what I have so far:

Im going to test for absolute convergence and if its fail then it would be conditional convergence.

pf:

as $(-1)^{n-1}$ alternating between -1 and 1 so

$\sum \vert a_n \vert$ = $\sum_{n=1}^{\infty}$ $\vert \frac{(-1)^{n-1}n}{(n+1)^2} \vert$ = $lim_{n \to \infty}$ $\frac{n}{(n^2+2n+1)}$ as n go to $\infty$ we have $\infty$ so the series is not absolute convergence therefore its conditional convergence.

any hint or anything I need to fix..thank you

tommy
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  • It is not true that $\frac{n}{n^2+2n+1}\to \infty$. And just because the series is not absolutely convergent doesn't mean it is conditionally convergent. It may be that it doesn't converge at all. – hmakholm left over Monica Dec 07 '14 at 00:24

2 Answers2

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Note that for $n > 1$,

$$\frac{n}{(n+1)^2} > \frac{n}{(n+n)^2} = \frac{1}{4n} $$

so the series is not absolutely convergent by the comparison test.

Since $n/(n+1)^2$ is monotonically decreasing to $0$ , the series converges conditionally by the alternating series test.

RRL
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  • so I cant say if not absolute convergence then it would be conditional convergence? then how would you test for conditional then – tommy Dec 07 '14 at 00:36
  • Correct. You check seperately for conditional convergence using for example the alternating series test. – RRL Dec 07 '14 at 00:37
  • ok if i use alternating series test wouldnt the series decreasing to 0? – tommy Dec 07 '14 at 00:39
  • The alternating series test says $\sum (-1)^n x_n$ converges to some finite value if $x_n > 0$ and $x_n$ converges as a monotonically decreasing sequence to $0$. The series itself does not necessarily converge to 0. – RRL Dec 07 '14 at 00:44
  • can you help me with one more problem – tommy Dec 07 '14 at 02:02
  • You are not yet eligible to chat, but you can either post the question or ask here if it is not too long. – RRL Dec 07 '14 at 02:26
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I think it will converge conditionally however since we have it will decrease and we can use the alternating series test to show that it will converge conditionally. I wanted to add that your limit is actually wrong just divide up and bottom by $n^2$ and take the limit.