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.

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    $\begingroup$ A good choice might be $a_n = \frac{1}{n^2}$. $\endgroup$ Apr 26, 2015 at 3:53
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    $\begingroup$ How did you come up with that? $\endgroup$
    – user234494
    Apr 26, 2015 at 3:55
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    $\begingroup$ @user234494 My logic was that $\frac{1}{3^{\sqrt{n}}}$ should decay very fast since it's a bit like a geometric series. As such, it should decay faster than $\frac{1}{n^2}$ since $\frac{1}{n^2}$ doesn't decay all that fast. $\endgroup$ Apr 26, 2015 at 3:59
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    $\begingroup$ As a way to prove (the other) inequality, you could write $n^2 = e^{2\ln n}$ and $3^{\sqrt{n}} = e^{\sqrt{n}\ln 3}$; and compare asymptotically the exponents. $\endgroup$
    – Clement C.
    Apr 26, 2015 at 4:03
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    $\begingroup$ Wouldn't you want to prove that $\dfrac{1}{3^{\sqrt n}} < \dfrac{1}{n^2}$ for large enough $n$ -- that is $n^2 < 3^{\sqrt{n}}$? $\endgroup$ Apr 26, 2015 at 4:04

5 Answers 5


Here's something different, that doesn't require comparing $n^2$ with $3^{\sqrt{n}}$, or any similar comparison:


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 $\log_e3 > 1$.

  • $\begingroup$ Yes. $a^{\log b} = b^{\log a}$ for $a$ and $b$ positive. $\endgroup$ Apr 26, 2015 at 5:29
  • $\begingroup$ Yes. I had initially thought that you didn't explicitly write this in your answer, but in fact it is there. $\endgroup$
    – shalop
    Apr 26, 2015 at 5:30
  • $\begingroup$ I suppose it might be "common knowledge" that $\sqrt{n}$ eventually exceeds $\log n$. But this is mathematically just as challenging to prove as the fact that $3^{\sqrt{n}}$ eventually exceeds $n^2$, the answer from the comment thread. $\endgroup$
    – 2'5 9'2
    Apr 26, 2015 at 5:51

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.

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    $\begingroup$ And how would convince someone that this integral is convergent? $\endgroup$
    – 2'5 9'2
    Apr 26, 2015 at 4:48
  • $\begingroup$ @alex.jordan: In general, for $n>0$ we have $\displaystyle\int_0^\infty e^{^{\LARGE-\sqrt[n]x}}~dx~=~n!~$ $\endgroup$
    – Lucian
    Apr 26, 2015 at 7:01
  • $\begingroup$ @alex.jordan, I would use substitution - $\sqrt x = t$. You can then solve the integral using integration by parts. $\endgroup$ Apr 26, 2015 at 8:02
  • $\begingroup$ @RoiBlumberg Oh good, then +1. Is it worth clarifying that in the answer? Maybe not if you are leaving something for OP to do. $\endgroup$
    – 2'5 9'2
    Apr 26, 2015 at 16:27
  • $\begingroup$ No, you're right, I will clarify that. $\endgroup$ Apr 27, 2015 at 16:08

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.)


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 $\sum_{n=1}^{\infty}3^{-\sqrt{n}}$.


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