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!
 A: Since $a^{1/\sqrt{n}}-1$ is asymptotic to $ \frac{\log a}{ \sqrt{n}},$ the series converges.
A: $$  3^t = e^{t \log 3} = 1 + t \log 3  + O(t^2) $$
$$  3^{1/\sqrt n} - 1 =   \frac{ \log 3}{\sqrt n} +  O(1/n) $$
$$  \frac{3^{1/\sqrt n} -1}{n}  =   \frac{ \log 3}{ n^{3/2}} +  O(1/n^2) $$
A: Using MVT $3^x-1=3^{\epsilon}\ln{3}\cdot (x-0),\epsilon \in (0,x)$ 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.
A: 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 $b > 0$.
