Question: Does the following series converge or diverge:

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

Many thanks!

Answer: 
$$ 3^t = e^{t \log 3} = 1 + t \log 3 + O(t^2) $$
$$ 3^{1/\sqrt n} - 1 = \frac{ \log 3}{\sqrt n} + O(1/n) $$
$$ \frac{3^{1/\sqrt n} -1}{n} = \frac{ \log 3}{ n^{3/2}} + O(1/n^2) $$
Using [MVT](https://en.wikipedia.org/wiki/Mean_value_theorem) $3^x-1=3^{\epsilon}\ln{3}\cdot (x-0),\epsilon \in (0,x)$ or
$$0<\frac{3^{\frac{1}{\sqrt{n}}}-1}{n}=3^{\epsilon}\ln{3}\frac{1}{n\sqrt{n}}<3\ln{3}\frac{1}{n\sqrt{n}}$$
because $\epsilon \in (0,\frac{1}{\sqrt{n}})$ or $0<\epsilon<1$ and $f(x)=3^x$ is ascending. As a result:
$$\sum\_{n=1}^{\infty}\frac{3^{\frac{1}{\sqrt{n}}}-1}{n}<3\ln{3}\sum\_{n=1}^{\infty}\frac{1}{n^{1+\frac{1}{2}}}$$
which [converges](https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#p-series).