Supposed to use the root and ratio test

[karush]

$\tiny{206.b.46}$
\begin{align*}
\displaystyle
S_{46}&=\sum_{k=1}^{\infty} \frac{2^k}{e^{k}-1 }\approx3.32569\\
% e^7 &=1+7+\frac{7^2}{2!}
%+\frac{7^3}{3!}+\frac{7^4}{4!}+\cdots \\
%e^7 &=1+7+\frac{49}{2}+\frac{343}{6}+\frac{2401}{24}+\cdots
\end{align*}
$\textsf{root test}$
$$\sqrt[k]{\frac{2^k}{e^{k}-1 } }
=2\left(\frac{1}{e^{k}-1 }\right)^{\frac{1}{k}}$$
$\textsf{this appears to converge }$

 

[Greg]

$$2\lim_{k\to\infty}\left(\frac{1}{e^k-1}\right)^{1/k}=\frac2e$$

Can you prove it?

 

[karush]

$$2\lim_{k\to\infty}\left(\frac{1}{e^k-1}\right)^{1/k}=\frac2e$$

Can you prove it?

$\text{not offhand but..this could be rewritten as}$
$$\displaystyle
2\lim_{{k}\to{\infty}}\sqrt[k]{\frac{1}{e^k-1} }$$

$\text{sorry spent about an hour looking for examples on this but not}(Sadface)