<s>[INST] 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.
 [/INST] 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. 
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