$\sum_{n=1}^\infty \frac{n^k}{1+2^n+(-2)^{nk}}$ $S = \sum_{n=1}^\infty \frac{n^k}{1+2^n+(-2)^{nk}}$
Find $k \in N: S$ coverages.
I started with checking the Cauchy's theorem for $k = 2m, m \in N$ and at this point everything is fine. But for $k = 2m - 1, m \in N$ things start to get complicated and I don't really have an idea how to proceed. 
 A: Let be
$$
a_n \left( k \right) = \frac{{n^k }}
{{1 + 2^n  + \left( { - {\text{2}}} \right)^{nk} }}
$$
If $k=1$ then the series is not convergent. Indeed, you have that
$$
a_n(1)  = \frac{n}
{{1 + 2^n  + \left( { - {\text{1}}} \right)^n 2^n }}
$$
Thus, when $n$ is odd you have that $a_n=n$. And the necessary condition for the convergence is not satisfied.
If $k$ is odd and greater than $1$ then your series is convergent. Indeed, 
$$
\begin{gathered}
  a_n \left( k \right) = \frac{{n^k }}
{{1 + 2^n  + \left[ {\left( { - {\text{1}}} \right)^k } \right]^n 2^{nk} }} =  \hfill \\
   \hfill \\
   = \frac{{n^k }}
{{1 + 2^n  + \left( { - {\text{1}}} \right)^n 2^{nk} }} =  \hfill \\
   \hfill \\
   = \frac{{n^k }}
{{\left( { - {\text{1}}} \right)^n 2^{nk} \left[ {1 + \frac{{1 + 2^n }}
{{\left( { - {\text{1}}} \right)^n 2^{nk} }}} \right]}} =  \hfill \\
   \hfill \\
   = \frac{{n^k \left( { - {\text{1}}} \right)^n }}
{{2^{nk} \left[ {1 + \frac{{\left( {1 + 2^n } \right)\left( { - {\text{1}}} \right)^n }}
{{2^{nk} }}} \right]}} \hfill \\ 
\end{gathered} 
$$
so that
$$
\begin{gathered}
  \left| {a_n \left( k \right)} \right| \leqslant \frac{{n^k }}
{{2^{nk} \left[ {1 - \frac{{\left( {1 + 2^n } \right)}}
{{2^{nk} }}} \right]}} \leqslant  \hfill \\
   \hfill \\
   \leqslant \frac{{n^k }}
{{2^{nk} \left[ {1 - \frac{1}
{2}} \right]}} = \frac{{2n^k }}
{{2^{nk} }} \hfill \\ 
\end{gathered} 
$$
which is true if $n$ is large enough. Now it is easy to prove that the series
$$
\sum\limits_{n = 1}^{ + \infty } {\frac{{2n^k }}
{{2^{nk} }}}  = 2\sum\limits_{n = 1}^{ + \infty } {\frac{{n^k }}
{{2^{nk} }}} 
$$
is convergent and so the given series is absolutely convergent.
