Does $\sum _{k=2} ^\infty \frac{(-1)^k}{\sqrt{k}+(-1)^k}$ converge conditionally? 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?
 A: Hint:
\begin{align*}
\frac{(-1)^n}{\sqrt{n}+(-1)^n} &= \frac{(-1)^n}{\sqrt{n}}\frac{1}{1+\frac{(-1)^n}{\sqrt{n}}} = \frac{(-1)^n}{\sqrt{n}}\left( 1-\frac{(-1)^n}{\sqrt{n}} + \frac{1}{n} + o\left(\frac{1}{n}\right) \right) \\
&=\frac{(-1)^n}{\sqrt{n}} -\frac{1}{n} + \left[ \frac{(-1)^n}{n^{3/2}} + o\left(\frac{1}{n^{3/2}}\right)  \right]
\end{align*}
The (series of the) first term converges conditionally, the second diverges, the third converges absolutely. What can you say about the sum of the three?
A: $$\sum _{k=2} ^\infty \frac{(-1)^k}{\sqrt{k}+(-1)^k}$$
multiply by the conjugate 
$$\sum _{k=2} ^\infty \frac{(-1)^k(\sqrt{k}-(-1)^k)}{(\sqrt{k}+(-1)^k)(\sqrt{k}-(-1)^k)}$$
$$\sum _{k=2} ^\infty \frac{(-1)^k\sqrt{k}-1}{k-1}$$
$$\sum _{k=2} ^\infty \frac{(-1)^k\sqrt{k}}{k-1}- \sum _{k=2} ^\infty\frac{1}{k-1}$$
The first sum converges by the alternating series test  but the second is a divergent harmonic sum .
A: 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.
