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?

Note that if $$k\le\sqrt{n}< k+1$$ then, since the function $x\mapsto 3^x$ is increasing, we have $$3^k\le3^{\sqrt{n}}<3^{k+1},\;\;\text{and}$$ $$\frac{1}{3^k}\ge\frac{1}{3^{\sqrt{n}}}>\frac{1}{3^{k+1}}$$for every $n\in [k^2,(k+1)^2[$ then 
\begin{align}
\sum_{n=1}^{N^2-1}{\frac{1}{3^{\sqrt{n}}}} & = \frac{1}{3^1}+\frac{1}{3^{\sqrt{2}}}+\frac{1}{3^{\sqrt{3}}}+\frac{1}{3^2}+\ldots+\underbrace{\frac{1}{3^{\sqrt{(N-1)^2}}}+\frac{1}{3^{\sqrt{(N-1)^2+1}}}+\ldots+\frac{1}{3^{\sqrt{N^2-1}}}}_{2N-1\,\text{terms}}\\
& \le \frac{1}{3^1}+\frac{1}{3^1}+\frac{1}{3^1}+\frac{1}{3^2}+\ldots+\underbrace{\frac{1}{3^{N-1}}+\ldots+\frac{1}{3^{N-1}}}_{2N-1\;\text{terms}}\\
& = \sum_{k=1}^{N-1}{\frac{2k+1}{3^k}}
\end{align}
Then, for comparison test, $\sum_{n=1}^{\infty}{\frac{1}{3^{\sqrt{n}}}}$ converges if $\sum_{k=1}^{\infty}{\frac{2k+1}{3^k}}$ does, and this happens due the ratio test.
