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?
 A: You can group entries according to $\lfloor \sqrt{n} \rfloor$ and bound the sum from above:
$$\sum_{n=1}^\infty \frac{1}{3^{\sqrt{n}}} = \sum_{k=1}^\infty \sum_{n=k^2}^{(k+1)^2-1} \frac{1}{3^{\sqrt{n}}} \le \sum_{k=1}^\infty \frac{2k+1}{3^k} = 2$$
Since your sequence consists of positive numbers only, this implies its convergence.
A: You can prove that: $3^{\sqrt{n}} > n^{\frac{3}{2}}$ for large enough $n$ and this inequality comes from: $3^x > x^3$ is true for all $x > 4$ especially integer value $x$. From this you can readily use the comparison test. In fact, you can make it a bit stronger.... because $3^x > x^k$ for a given positive $k$
A: 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.
A: 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 $3^\sqrt{x} = e^{\left(\sqrt{x} \ln\left(3\right)\right)}$
