For which $\alpha$ does the series $\sum_{n = 1}^{\infty}\big(2^{n^{-\alpha}}\!\! - \!1\big)$ converge? For what $\alpha$,  does $\displaystyle\sum_{n = 1}^{\infty}\big(2^{n^{-\alpha}} - 1\big)$ converge?
Divergence of $\sum_{n = 1}^{\infty}(2^\frac{1}{n} - 1)$ prompted this question.
 A: The answer is that the series converges if and only if $a>1$.
Using the Mean Value Theorem for $f(x)=\exp(x\log 2)$ at $x=0$ we obtain
$$
2^{n^{-a}}=1+\frac{\log 2}{n^a}\mathrm{e}^{\vartheta n^{-a}\log 2 }
$$
for some $\vartheta\in (0,1)$,
and hence
$$
\frac{\log 2}{n^a}<2^{n^{-a}}-1<2\cdot \frac{\log 2}{n^a}.
$$
Hence the series
$$
\sum_{n=1}^\infty(2^{n^{-a}}-1),
$$
converges if and only if the series 
$$
\sum_{n=1}^\infty \frac{1}{n^a},
$$
converges, i.e., if and only if $a>1$.
A: We write $f(\alpha) = \sum_{n\geq 1} (2^{1/n^{\alpha}} -1)$. Obviously, $f$ decreases with $\alpha$. Moreover, you know that $f(1) = \infty$. Let fix $\alpha>1$. We have
\begin{eqnarray}
 f(\alpha) &=& \sum_{n\geq 1} (e^{\log(2)/n^{\alpha}} -1) \\
&=& \sum_{n\geq 1} \sum_{k\geq 1} \frac{1}{k!}\frac{\log(2)^k}{n ^{\alpha k}} \\
&=& \sum_{k\geq 1} \frac{\log(2)^k}{k!} \sum_{n\geq 1} \frac{1}{n^{\alpha k}} \\
&=& \sum_{k\geq 1} \frac{\log(2)^k}{k!} \zeta (\alpha k).
\end{eqnarray}
Using that $\zeta(\alpha k) \leq \zeta (\alpha)$ for $k\geq 1$, you finally have $$f(\alpha) \leq \sum_{k\geq 1} \frac{\log(2)^k}{k!} \zeta(\alpha) = \zeta(\alpha) < \infty.$$
The serie converges iff. $\alpha >1$. 
