TITLE: Does $\sum _{k=2} ^\infty \frac{(-1)^k}{\sqrt{k}+(-1)^k}$ converge conditionally?
QUESTION [7 upvotes]: Discuss the convergence of the sum: 
$$\sum _{k=2} ^\infty \frac{(-1)^k}{\sqrt{k}+(-1)^k}.$$
My answer so far: It does not converge absolutely since 
$$\left | \frac{(-1)^k}{\sqrt{k}+(-1)^k} \right | \geq \frac {1} {\sqrt{k} + 1},$$
the sum of whose terms diverges. Intuitively, I think it should converge conditionally because $ \sum \frac{(-1)^k}{\sqrt{k}}$ does, and the $(-1)^k$ in the denominator should not make a big difference. However, I am having trouble demonstrating its conditional convergence.
Evidently, the problem is that $\frac{1}{\sqrt{k}+(-1)^k}$ is not monotone. Let's call this sequence $a_n$. If we can show that 
$$\sum_{k=2}^\infty |a_k - a_{k-1}| < \infty,$$ 
then we're done by Dirichlet's test. But this seems to be $O(1/k)$. 
Any ideas?

REPLY [6 votes]: Let $k$ be even, and add the $k$-th term and the $(k+1)$-th. We get 
$$\frac{\sqrt{k+1}-\sqrt{k}-2}{(\sqrt{k}+1)(\sqrt{k+1}-1)}.$$
It is not hard to show that $\sqrt{k+1}-\sqrt{k}$ approaches $0$ as $k\to\infty$. So after a while the top is negative and has absolute value $\gt 1$.
The denominator is after a while less than $2k$.
So after a while the sums of two consecutive terms is negative and has absolute value $\gt \frac{1}{2k}$. It follows that our series diverges.