# Does this series converge? Alternating series and ratio tests do not work

Does the sum $$\sum_{n=1}^\infty (-1)^n\frac{n}{n+2}.$$ converge?

The conditions for the Leibniz alternating series test are not satisfied, as $$\frac{n}{n+2}\nrightarrow 0$$ as $$n\rightarrow\infty$$. Also, the ratio test yields an answer of $$R=1$$ so is inconclusive. Any ideas how to do it?

• Hint. You already noted that the general term does not have limit $0$. Think a bit before you start on tests. Apr 1 at 17:47
• Alongside the HINT left by Ethan, note that $\frac{n}{n+2}=1-\frac{2}{n+2}$. Apr 1 at 17:49

No, that series does not converge, as the sequence $$x_n = (-1)^n\frac{n}{n+2}$$ does not converge to $$0$$. It does not even converge. Probably, the most simple way to see this is to consider the subsequences $$a_n:=x_{2n} = \frac{2n}{2n+2}$$ and $$b_n:=x_{2n+1} = -\frac{2n+1}{2n+3}.$$ Obviously, $$a_n\to 1$$ and $$b_n\to-1$$ as $$n\to\infty$$, hence $$\limsup_{n\to\infty}x_n\neq\liminf_{n\to\infty}x_n.$$