Question: Does the following series converge or diverge:

$$\sum\_{n=1}^\infty \frac{3^{1/\sqrt{n}}-1}{n}$$

Many thanks!

Answer: 
$$ 3^t = e^{t \log 3} = 1 + t \log 3 + O(t^2) $$
$$ 3^{1/\sqrt n} - 1 = \frac{ \log 3}{\sqrt n} + O(1/n) $$
$$ \frac{3^{1/\sqrt n} -1}{n} = \frac{ \log 3}{ n^{3/2}} + O(1/n^2) $$
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 $b > 0$.