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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.
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.
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 $\log_e3 > 1$.
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.
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.)
