Does $\sum 3^{-\sqrt{n}}$ converge or diverge? I need to find out whether this series converges or diverges:
$$\sum_{n=1}^\infty \frac 1{3^{\sqrt{n}}}$$
The $n$th term, ratio, and root tests are inconclusive, Abel's test doesn't apply (or I can't think of how to separate out part of the sequence), and I can't figure out a series to compare this to that'll work.
WolframAlpha says it converges by the comparison test, BTW.  It just doesn't tell me what it compared the series to.
 A: Here's something different, that doesn't require comparing $n^2$ with $3^{\sqrt{n}}$, or any similar comparison:
$$\frac{1}{3^{\sqrt{n}}}\leq\frac{1}{3^{\left\lfloor\sqrt{n}\right\rfloor}}$$
And the sequence $\{\left\lfloor\sqrt{1}\right\rfloor,\left\lfloor\sqrt{2}\right\rfloor,\left\lfloor\sqrt{3}\right\rfloor,\ldots\}$ is equal to $$\{\overbrace{1,1,1}^3,\overbrace{2,\ldots,2}^{5},\overbrace{3,\ldots,3}^{7},\ldots,\overbrace{k,\ldots,k}^{2k+1},\ldots\}\text{.}$$ This follows from understanding that consecutive perfect squares differ by increasing odd numbers. Or equivalently that the sum of consecutive odd integers $3+5+7+\cdots$ is always $1$ shy of a perfect square.
So
$$\begin{align}
\sum_{n=1}^{\infty}\frac{1}{3^{\sqrt{n}}}&<\sum_{n=1}^{\infty}\frac{1}{3^{\left\lfloor\sqrt{n}\right\rfloor}}\\
&=\sum_{k=1}^{\infty}\frac{2k+1}{3^{k}}\\
&=2\sum_{k=1}^{\infty}\frac{k}{3^{k}}+\sum_{k=1}^{\infty}\frac{1}{3^{k}}\\
&=2\cdot\frac{3}{4}+\frac{1}{2}\\
&=2
\end{align}$$
Not only is the sum convergent, it's less than $2$. You can get a better upper bound by leaving the initial terms alone instead of using the floor function. For instance, this same approach can be used with $$\begin{align}\frac{1}{3^{\sqrt{1}}}+\frac{1}{3^{\sqrt{2}}}+\frac{1}{3^{\sqrt{3}}}+\sum_{n=4}^{\infty}\frac{1}{3^{\left\lfloor\sqrt{n}\right\rfloor}}
&=\frac{1}{3^{\sqrt{1}}}+\frac{1}{3^{\sqrt{2}}}+\frac{1}{3^{\sqrt{3}}}+1\\
&\approx1.69\ldots
\end{align}$$
which is a better upper bound. (A CAS says the true value is approximately $1.34\ldots$)

For an even better approximation that you can't immediately tell is over or under, replace each term in $\sum_{n=N^2}^{(N+1)^2-1}\frac{1}{3^{\sqrt{n}}}$ (the portion of the series corresponding to one of the constant substrings in $\{\left\lfloor\sqrt{n}\right\rfloor\}$) with the average of the end terms: $\frac{1}{3^{\sqrt{N^2}}}$ and $\frac{1}{3^{\sqrt{(N+1)^2}}}$. So 
$$\begin{align}
\sum_{n=1}^{\infty}\frac{1}{3^{\sqrt{n}}}
&\approx\sum_{N=1}^{\infty}\frac12\left(\frac{1}{3^N}+\frac{1}{3^{N+1}}\right)(2N+1)\\
&=\frac43
\end{align}$$
I'm not offering error analysis, but you can note that this does indeed get within $0.6\%$ of the exact value. 
A: Hint: $\log_e n \leq \sqrt{n}$ for $n\geq 1$. Then
$$3^{\log_e n} = n^{\log_e3}\leq 3^{\sqrt{n}}$$
This will imply, for n large enough
$$\frac{1}{n^{\log_e3}}\geq \frac{1}{3^{\sqrt{n}}}$$
Now $\log_e3 > 1$.
A: One way to show that this series converges can be to compare it to $\int_0^\infty (1/3)^\sqrt x dx$, which is convergent.
Edit - You can solve this integral by substituting $t=\sqrt x$, and then integrating the resulting integral by parts.
A: If we stick to general tests, we
can apply the condensation test ($a_n=\frac{1}{3^{\sqrt{n}}}$ is decreasing) + the $n$-th root (for example) test.
$$
\sum_n a_n \le \sum_n (2n+1)a_{n^2}=\sum_n \frac{2n+1}{3^n}\\
\lim_{n} \left(\frac{2n+1}{3^n}\right)^{\frac{1}{n}}=\frac{1}{3}<1
$$
(The condensation is done through the $\varphi(n)=n^2$ subsequence, in this way we "eliminate" the square root.)
A: Or you can apply https://en.wikipedia.org/wiki/Integral_test_for_convergence
Here, $f(x)=3^{-\sqrt{x}}$, then your series converges if and only if $\int_1^{\infty}3^{-\sqrt{x}}dx$ converges.
But $\int_1^{\infty}3^{-\sqrt{x}}dx\overset{x=u^2}=\int_1^{\infty}2ue^{-(\ln3)u}du=-\frac{2u}{\ln3}e^{-(\ln3)u}\vert_1^{\infty}+\frac{2}{\ln3}\int_1^{\infty}e^{-(\ln3)u}du=\frac{2}{\ln3}+\frac{2}{(\ln3)^2}$, it converges so does the series $\sum_{n=1}^{\infty}3^{-\sqrt{n}}$.
