Is $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent? Is   $\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}}$ convergent ?
I use it to compare with $1/n^2$, and then I used LHôpitals rule multiple times. Finally , I can solve it. However,I think we have other methods!  somebody can help me?

By using the integral test:
The function $\frac{1}{3^{\sqrt{x}}}$ is continuous, positive and decreasing on $[1,\infty)$. We then evaluate the integral:
\begin{align}
\int_1^\infty\, \frac{1}{3^{\sqrt{x}}}\, dx = \frac{2}{3 \, \log\left(3\right)} + \frac{2}{3 \, \log\left(3\right)^{2}} \approx 1.15918311754471
\end{align}
and see that it has a finite value, which in turn proves that the sum is also finite.
It can also be used to get the lower and upper bounds:
\begin{align}
\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}} &\ge \int_1^\infty\, \frac{1}{3^{\sqrt{x}}}\, dx\\
&\ge \frac{2}{3 \, \log\left(3\right)} + \frac{2}{3 \, \log\left(3\right)^{2}} \approx 1.15918311754471
\end{align}
and 
\begin{align}
\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}} &\le \frac{1}{3}+\int_1^\infty\, \frac{1}{3^{\sqrt{x}}}\, dx\\
&\le \frac{1}{3}+\frac{2}{3 \, \log\left(3\right)} + \frac{2}{3 \, \log\left(3\right)^{2}} \approx 1.49251645087804
\end{align}
The actual numerical result using a program:
\begin{align}
\sum_{n=1}^\infty {1\over 3^{\sqrt{n}}} \approx 1.34070489742952420178654874099073605072839417287611074076358138561327
\end{align}
p.s. the integral can be easily solved by observing that $3^\sqrt{x} = e^{\left(\sqrt{x} \ln\left(3\right)\right)}$
