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First I tried using the ratio test but that did not work because it was inconclusive. I think I have to use the alternating series test. If I can prove that the underlying sequence diverges then I can then say that the series diverges, however I do not know where to begin to show that the sequence $(-1)^n(n)/(n+2)$ diverges. Any tips on what methods to use would be appreciated. Thanks.

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The nth-term test says that if $\displaystyle\lim_{n\to\infty} a_n\neq 0$ or this limits does not exist, then $\sum_{n=1}^\infty a_n$ diverges. In your case one can factor out $n$ of the absolute value and see that $$\frac{n}{n+2}=\frac{1}{1+2/n}\overset{n\to\infty}{\longrightarrow}1\neq 0.$$ Hence $$\sum_{n=1}^\infty\frac{n}{n+2}$$ diverges and thus $$\sum_{n=1}^\infty(-1)^n\frac{n}{n+2}$$ as well (in accordance to the alternating series test).

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  • $\begingroup$ Why is n/(n+2) the absolute value? $\endgroup$ – gman9732 Mar 26 '16 at 19:47
  • $\begingroup$ never mind the absolute value of (-1)^n is 1 :) $\endgroup$ – gman9732 Mar 26 '16 at 20:07
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A necessary condition for a series to converge is that the terms converge to zero. Clearly $\frac{-1^n(n)}{n+2}$ does not converge to zero.

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  • $\begingroup$ I can tell by looking at it that it does not converge to 0, but how would you show that mathematically? $\endgroup$ – gman9732 Mar 26 '16 at 19:17
  • $\begingroup$ its absolute value is always larger than $\frac{1}{3}$. $\endgroup$ – Jorge Fernández Hidalgo Mar 26 '16 at 19:20
  • $\begingroup$ @Garrett $$\frac{n}{n+2}=\frac{1}{1+2/n}\overset{n\to\infty}{\longrightarrow}1\neq 0$$ $\endgroup$ – Christian Ivicevic Mar 26 '16 at 19:29
  • $\begingroup$ @Christian I understand that, but what about the (-1)^n, why does that not matter? $\endgroup$ – gman9732 Mar 26 '16 at 19:32
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    $\begingroup$ A sequence converges to zero if and only if its absolute value converges to zero. $\endgroup$ – Jorge Fernández Hidalgo Mar 26 '16 at 19:33
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The absolute value of the summand does not tend to 0, hence series diverges

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  • $\begingroup$ Could you explain what you mean by "summand"? $\endgroup$ – gman9732 Mar 26 '16 at 19:15
  • $\begingroup$ I mean $|a_n|$, it does not tend to 0 $\endgroup$ – Alex Mar 26 '16 at 19:16

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