# $\sum_{n=1}^\infty \frac{n^k}{1+2^n+(-2)^{nk}}$

$$S = \sum_{n=1}^\infty \frac{n^k}{1+2^n+(-2)^{nk}}$$

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.

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.