series involving $(-1)^k$ Given the series ${a_n}$ : 
$a_n= \sum _{k=1}^{n}{\frac { \left( -1 \right) ^{k}}{{3}^{k}+\sqrt {k}}}$
prove that ${a_n}$ converges. 
So far this is what I've done:
I've split this into 2 sub-series: $k=2n$ and $k=2n+1$
for $(k=2n)$ I get: 
$\sum _{k=1}^{n} \left( {3}^{k}+\sqrt {k} \right) ^{-1} = 0$
(as k goes to infinity the whole series converges to 0)
for $(k=2n+1)$ I get: 
$\sum _{k=1}^{n}- \left(  \left( {3}^{k} \right) ^{-1}+\sqrt {-k}
 \right) ^{-1}$ 
(which also converges to 0 as k approaches to infinity)
Since both subs-series of $a_n$ converge to the same value, $0$, $a_n$ converges, in particular to $0$. 
I'm not sure if this is a tight enough proof since maybe by chance it works out here specifically. I also checked separate limits as k approaches $-\infty$. They both converge to $0$ for each sub-series. 
What am I missing?
Thanks. 
 A: $$
|a_n| \le \sum_{k=1}^n \frac{ 1}{3^k + \sqrt k} \le \sum_{k=1}^n \frac{1}{3^k - \frac 12 3^k} < 2 \sum_{k=1}^n \left( \frac 13 \right)^k
$$
which converges because this is a geometric series. Since $a_n$ converges absolutely it converges conditionally.
You could also use Leibniz's criterion, which states that if a series has leading term $(-1)^k b_k$ with $b_k \to 0$, then $\sum (-1)^k b_k$ converges. The fact that $b_k = \frac 1{3^k + \sqrt k} \to 0$ is clear enough.
EDIT : As the comments pointed out, I didn't remember at the time all the hypothesis of Leibniz's Criterion, which are that on top of what I said, the $b_k$'s must be, up to some $N$, non-negative and decreasing.
Hope that helps,
A: You can use Leibniz's Rule for alternating series.
If $\{ a_n\}$ is a monotonic decrasing sucesion of numbers, the alternate series converges, and if $S$ is the sum, then, for every $n > 0$: $$ 0 < (-1)^n (S-s_n) < a_{n+1}$$ and $$0 < S < a_1$$
Thus your series converges, and you can calculate it's sum approximately with the first inequality.
