8
$\begingroup$

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.

$\endgroup$
9
  • 5
    $\begingroup$ A good choice might be $a_n = \frac{1}{n^2}$. $\endgroup$ Apr 26, 2015 at 3:53
  • 1
    $\begingroup$ How did you come up with that? $\endgroup$
    – user234494
    Apr 26, 2015 at 3:55
  • $\begingroup$ It's one of the "simplest converging series," besides the geometric ones. Clearly, yours is not geometric, so among those who remain $\sum_n \frac{1}{n^2}$ would be a natural candidate. $\endgroup$
    – Clement C.
    Apr 26, 2015 at 3:56
  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ 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

4 Answers 4

9
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $\log_e3 > 1$.

$\endgroup$
3
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ 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
1
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.