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?
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.
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?
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} $.
