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

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.

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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.

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