Prove the series $\sum_{n=1}^∞ (-1)^n(n)/(n+2)$ diverges

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.

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).

• Why is n/(n+2) the absolute value? – gman9732 Mar 26 '16 at 19:47
• never mind the absolute value of (-1)^n is 1 :) – gman9732 Mar 26 '16 at 20:07

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.

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

The absolute value of the summand does not tend to 0, hence series diverges

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