Convergence and absolute convergence of $\sum_{n=1}^{\infty} = {(-1)^n \over n + (-1)^{n-1}}$ I am trying to conclude about the convergence and absolute convergence of
$$\sum_{n=1}^{\infty} = {(-1)^n \over n + (-1)^{n-1}}$$
For absolute convergence, we can note that
$$\lvert a_n \rvert = {1 \over n + (-1)^{n-1}}$$
$$\sum_{n=1}^{\infty} = {1 \over n + (-1)^{n-1}} = {1 \over 2} + 1 + {1 \over 4} + {1 \over 3} + {1 \over 6} + {1 \over 5} + \dots$$
We can see that this is a harmonic series with the terms rearranged. The sequence of partial sums will be strictly monotonic and for even numbers the terms will be equal to the terms of the sequence of partial sums of the harmonic series. This means that the series doesn't converge, so we have no absolute convergence.
Now, how can we conclude about convergence? ${1 \over n + (-1)^{n-1}}$ is not monotonic, so the tests I have covered so far (Leibniz, Dirichlet and Abel) are not applicable.
 A: If we just study the $2N$-th partial sum
$$\sum_{n=1}^{2N}\frac{(-1)^n}{n+(-1)^{n-1}} = \sum_{k=1}^{N}\frac{1}{2k-1}-\sum_{k=1}^{N}\frac{1}{2k}=\sum_{k=1}^{N}\frac{1}{2k(2k-1)} $$
we trivially have that our series is conditionally convergent and
$$ \sum_{n\geq 1}\frac{(-1)^n}{n+(-1)^{n-1}} = \color{blue}{\log 2}.$$
A: Note that $$\sum_{n\leq2N}\frac{\left(-1\right)^{n}}{n+\left(-1\right)^{n-1}}=\sum_{n\leq N}\frac{1}{2n-1}-\sum_{n\leq N}\frac{1}{2n}=\sum_{n\leq2N}\frac{\left(-1\right)^{n+1}}{n}
 $$ hence $$\sum_{n\geq1}\frac{\left(-1\right)^{n}}{n+\left(-1\right)^{n-1}}=\sum_{n\geq1}\frac{\left(-1\right)^{n+1}}{n}=\log\left(2\right).
 $$ For the other case we have $$\sum_{n\leq2N}\frac{1}{n+\left(-1\right)^{n-1}}=\sum_{n\leq N}\frac{1}{2n-1}+\sum_{n\leq N}\frac{1}{2n}=\sum_{n\leq2N}\frac{1}{n}$$ hence the series diverges.
A: In fact, we can evaluate the series in closed form.   Proceeding, we write 
$$\begin{align}
\sum_{n=1}^{2N}\frac{(-1)^{n}}{(-1)^{n-1}+n}&=\sum_{n=1}^N\left(\frac{1}{2n-1}-\frac{1}{2n}\right)\\\\&=\sum_{n=1}^N\left(\frac{1}{2n-1}+\frac{1}{2n}\right)-\sum_{n=1}^N\frac1n\\\\&=\sum_{n=1}^{2N}\frac1n-\sum_{n=1}^N\frac1n
\\\\&=\sum_{n=1}^{N}\frac{1}{n+N}\\\\
&=\frac1N\sum_{n=1}^{N}\frac{1}{1+n/N} \tag {1}\\\\&\to \int_0^1 \frac{1}{1+x}\,dx\,\,\text{as}\,\,N\to \infty \tag{2}\\\\
&=\log(2)
\end{align}$$
where we used only elementary arithmetic to take us to $(1)$ and recognized the sum in $(1)$ as a Riemann sum to arrive at $(2)$.  
An alternative way forward to evaluating the series is to write $$\sum_{n=1}^N\left(\frac{1}{2n+1}-\frac{1}{2n}\right)=\sum_{n=1}^{2N}\frac{(-1)^{n-1}}{n}$$
Then, recalling that $\log(1+x)$ has Taylor series representation $\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}$ for $-1<x\le 1$, we see that $$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(-1)^n+n}=\log(2)$$as expected!
