Let $$f(k) = \dfrac{\sum_{n=1}^\infty 2^n e^{-\frac{2^n}{nk}}}{k \log k}$$ where $k \geq 2$ is an integer. From numerical investigation, and from reasons stemming from this question, I believe that there is a positive constant $B < \infty$ such that $$|f(k)| < B$$ for all $k.$
My question is:
How do I prove that there exists such a constant?
We have for $\eta > 0$
$$ \eta^2 e^{-\eta} \lt 1 \Rightarrow \eta e^{-\eta}\lt \frac{1}{\eta} $$
then
$$ \sum_{\nu = 1}^{\infty}\nu\left(\frac{2^{\nu}}{\nu k}\right)e^{-\frac{2^{\nu}}{\nu k}} \lt \sum_{\nu=1}^{\infty}\nu\left(\frac{\nu k}{2^{\nu}}\right) = k\sum_{\nu=1}^{\infty}\frac{\nu^2}{2^{\nu}} $$
We know that
$$ \sum_{\nu = 1}^{\infty}\nu^2x^{\nu} \lt M \ \ \mbox{for } |x| < 1 $$
For $x = 2^{-1}$ we have $M = 6$ So we can conclude that
$$ f(k) \lt \frac{N}{\log (k)} $$