Series Convergence / Divergence: $\sum \frac{3^{1/\sqrt{n}}-1}{n}$ Does the following series converge or diverge:
$$\sum_{n=1}^\infty \frac{3^{1/\sqrt{n}}-1}{n}$$
Many thanks!

For any $a > 1$ and $b > 0$,
consider
$\sum_{n=1}^\infty \frac{a^{1/n^b}-1}{n}
$.
I will show that this sum converges.
OP's problem is
$a=3$ and $b = \frac12$.
$a^{1/n^b}
=e^{\ln a/n^b}
$.
If $0 < x < \frac12$,
$\begin{array}\\
e^x
&=\sum_{k=0}^{\infty} \frac{x^k}{k!}\\
&<\sum_{k=0}^{\infty} x^k\\
&=\frac1{1-x}\\
&< 1+2x
\qquad\text{since } (1-x)(1+2x) = 1+x-2x^2
= 1+x(1-2x) > 1\\
\end{array}
$
Therefore,
for
$\ln a/n^b < \frac12$
(i.e.,
$n > (2\ln a)^{1/b}=N(a, b)$),
$e^{\ln a/n^b}
\lt 1+2\ln a/n^b
$
so that,
for $n > N(a, b)$,
$a^{1/n^b}-1
\lt \frac{2\ln a}{n^b}
$.
Therefore
$\frac{a^{1/n^b}-1}{n}
\lt \frac{2\ln a}{n^{b+1}}
$
and $\sum_{n > N(a, b)} \frac{2\ln a}{n^{b+1}}
$
converges since $b > 0$.
