Test the convergence of $\sum _{k=1}^\infty (-1)^{k-1} \frac{2^k}{2^k+k^2}$ Test the convergence of $\sum _{k=1}^\infty (-1)^{k-1} \frac{2^k}{2^k+k^2}$
my idea
$a_k=\sum _{k=1}^\infty (-1)^{k-1} \frac{2^k}{2^k+k^2}$
$a_{k+1}=\sum _{k=1}^\infty (-1)^{k} \frac{2^{k+1}}{2^{k+1}+(k+1)^2}$
$|\frac{a_{k}}{a_{k+1}}|_{k\to \infty}=\left| \frac{1}{2}\frac{2^{k+1}+(k+1)^2}{2^k+k^2}\right|_{k\to\infty}$
can any help from here please thank you
 A: If the series $\sum_{k>0}a_k$ converges then $a_k\to 0$. So if $\lim_{k\to \infty} a_k \neq 0$ or the limit does not exist then your series diverges. Now calculate:
\begin{align} 
\lim_{k\to \infty} a_k = \lim_{k\to \infty} \frac{(-1)^{k-1}2^k}{2^k+k^2} =....
\end{align} 
This limit does not exist, can you conclude? 
Here is used: "P implies Q" is  equivalent with "not Q implies not P". I hope this helps. 
A: Well, 
\begin{align*}
\sum_{k=1}^{\infty} \frac{(-1)^{k-1} 2^k}{2^k+k^2} &= \sum_{k=1}^{\infty} (-1)^{k-1} \frac{2^k+k^2-k^2}{2^k+k^2} \\ 
 &= \sum_{k=1}^{\infty} (-1)^{k-1} \left [ 1 - \frac{k^2}{2^k+k^2} \right ]\\ 
 &= \sum_{k=1}^{\infty} (-1)^{k-1} - \sum_{k=1}^{\infty} \frac{(-1)^{k-1} k^2}{2^k+k^2}
\end{align*}
The first series obviously diverges whereas the second converges. To see that split it apart. For example
$$\sum_{k=1}^{\infty} \frac{(-1)^{k-1} k^2}{2^k+k^2} = \sum_{k=1}^{\left \lfloor \frac{2}{\log 2} \right \rfloor} \frac{(-1)^{k-1} k^2}{k^2+2^k} + \sum_{\left \lfloor \frac{2}{\log 2} \right \rfloor+1}^{\infty} \frac{(-1)^{k-1} k^2}{2^k+k^2} $$
The first sum is finite , hence convergent , and the second converges since $\displaystyle \lim_{k\rightarrow +\infty} \frac{k^2}{2^k+k^2} = 0 $ and of course the sequence $\displaystyle a_k = \frac{k^2}{2^k+k^2}$ is decreasing. Here is the invoked test.  
Thus your sum diverges. 
