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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?

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    $\begingroup$ Hint. You already noted that the general term does not have limit $0$. Think a bit before you start on tests. $\endgroup$ Apr 1 at 17:47
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    $\begingroup$ Alongside the HINT left by Ethan, note that $\frac{n}{n+2}=1-\frac{2}{n+2}$. $\endgroup$
    – Mark Viola
    Apr 1 at 17:49
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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 $$\limsup_{n\to\infty}x_n\neq\liminf_{n\to\infty}x_n.$$

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