Find if $\sum_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)}$ converges Find if the following series is convergent or divergent, justify.
$$\sum_{n=1}^\infty  \frac{(-1)^n}{n(2+(-1)^n)}$$
My first idea was to use absolute convergence to get rid of both $(-1)^n$, take $1/2$ out to be left with the harmonic series but I don't think the absolute value will get rid of the $(-1)^n$ in the denominator. 
Where do I go from there? 
 A: Since
\begin{align}
\sum_{n=1}^\infty  \frac{(-1)^n}{n(2+(-1)^n)} &= \sum_{n=1}^{\infty} \frac{(-1)^n (2 - (-1)^n)}{n \, (2 + (-1)^{n})(2 - (-1)^n)} \\
&= \frac{1}{3} \, \sum_{n=1}^{\infty} \left( \frac{2 \, (-1)^n}{n} - \frac{1}{n} \right) \\
&= - \frac{2}{3} \, \ln(2) - \lim_{n \to \infty} H_{n},
\end{align}
where $H_{n}$ is the Harmonic number. Since the Harmonic number diverges as $n \to \infty$ then the series diverges.
A: Hint: Writing out the first few terms, we get
$$-\frac{1}{1}+\frac{1}{6}-\frac{1}{3}+\frac{1}{12}-\frac{1}{5}+\frac{1}{18}-\frac{1}{7}+\frac{1}{24}-...$$
Can you find a comparison to use to simplify the relations between the terms to show that the negative terms are "too large" for the series to converge?
A: The terms go to zero
and the sum of the
first $2m$ terms is
$\begin{array}\\
\sum_{n=1}^{2m}  \frac{(-1)^n}{n(2+(-1)^n)}
&=\sum_{n=1}^{m}  \left(\frac{(-1)^{2n-1}}{(2n-1)(2+(-1)^{2n-1})}+\frac{(-1)^{2n}}{(2n)(2+(-1)^{2n})}\right)\\
&=\sum_{n=1}^{m}  \left(\frac{-1}{(2n-1)(2-1)}+\frac{1}{(2n)(2+1)}\right)\\
&=\sum_{n=1}^{m}  \left(\frac{-1}{2n-1}+\frac{1}{6n}\right)\\
&=\sum_{n=1}^{m}  \left(\frac{(-6n)+(2n-1)}{(2n-1)(6n)}\right)\\
&=-\sum_{n=1}^{m}  \frac{4n+1}{(2n-1)(6n)}\\
\end{array}
$
and the sum of this diverges since
$\frac{4n+1}{(2n-1)(6n)}
\gt \frac{4n}{(2n)(6n)}
= \frac{1}{3n}
$.
