Convergence of $\sum2^{-\sqrt{k}} $

Question

So I am going to determine whether this series converges or not: $$\sum_{k=0}^\infty 2^{-\sqrt{k}} $$

Since this chapter is about the ratio test, I applied that test to this series. I end up with this limit $$\lim_{k \to \infty} 2^{\sqrt{k}-\sqrt{k+1}}$$

I'm stuck here, don't know how to calculate this limit. I could simplify to: $$\sqrt{k}-\sqrt{k+1} = \frac{1}{\sqrt{k}+\sqrt{k+1}} $$ But I doubt this will help me.

Could anyone help me?

Okay so the ratio test doesn't work then? Which test could I use then? — fejz1234, Sep 03 '16 at 13:02
ratio and root test are inconclusive. You will need to compare — operatorerror, Sep 03 '16 at 13:10
@fejz1234 I have answered. Although I really like the integral test if you are familiar with it; I didn't see it — operatorerror, Sep 03 '16 at 13:32

User8128 · Answer 1 · 2016-09-03T14:30:41.330

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The integral test works. We calculate \begin{align*} \int^\infty_0 2^{-\sqrt x} &= \int^\infty_0 e^{-\sqrt{x}\log 2} dx \\ &= \frac{ 2 }{(\log 2)^2}\int^\infty_0 te^{-t} dt \,\,\,\,\,\,\, [t = \sqrt x \log 2]. \end{align*} The latter converges (easily shown via integration by parts) so the sum converges as well.

EDIT: As suggested in the above comments, a comparison should work as well. For sufficiently large $k$, we see $\sqrt{k} \ge 2\log_2 k$. This follows since asymptotically $\sqrt{k} \gg \log_2 k$. For such $k$, we have $$2^{-\sqrt k} \le 2^{-2\log_2 k} = \frac 1 {k^2}.$$

edited Sep 03 '16 at 14:30

answered Sep 03 '16 at 13:17

User8128

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This is really the simplest way, for sure. – Claude Leibovici Sep 03 '16 at 13:29
@ClaudeLeibovici Does mine not work? I do really like this solution – operatorerror Sep 03 '16 at 13:30
@qbert. For sure, it works ! – Claude Leibovici Sep 03 '16 at 13:39
I didnt really follow on the the integral-step. After the third equality, you seem to break out 2(log2)^2 and whats inside the integral seems to have changed? – fejz1234 Sep 03 '16 at 13:41
Could anyone help me with this step: $$ \int^\infty_0 e^{-\sqrt{x}\log 2} dx $$ $$= 2 (\log 2)^2\int^\infty_0 \sqrt{x}\cdot{log2}\cdot{e^{-\sqrt{x}\cdot{log2}}} dx$$ I cant see how this holds. – fejz1234 Sep 03 '16 at 14:09
Just make the substitution: $t = \sqrt x \log 2$ gives $$dt =\frac {dx}{2\sqrt x} \log 2 = \frac{dx}{2\frac t {\log2}} \log2$$ so $$dx = \frac {2t}{(\log 2)^2}dt$$ so it seems I made a minor mistake with the constant (I'll edit). The rest is fine. You can't just replace the $t$ in that line with $\sqrt x\log 2$ because you also need to replace the $dt$. – User8128 Sep 03 '16 at 14:28
Thanks very much! Got it now. – fejz1234 Sep 03 '16 at 14:41
Integral test, great. But for heaven's sake don't say that the sum is equal to the integral... – David C. Ullrich Sep 03 '16 at 16:28
Agreed. As a calculus TA, I saw that all too often. – User8128 Sep 03 '16 at 16:42

score 5 · Answer 2 · answered Sep 03 '16 at 13:25

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It is not difficult to prove, if $k $ is sufficiently large, $k\geq K $ say, that $$\sqrt{2^{k}}\geq2k $$so if we use the Cauchy condensation test we note that $$\sum_{k\geq K}\frac{2^{k}}{2^{\sqrt{2^{k}}}}\leq\sum_{k\geq K}\left(\frac{1}{2}\right)^{k}$$ and so we can conclude that the series converges.

answered Sep 03 '16 at 13:25

Marco Cantarini

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Why the downvote??? – Marco Cantarini Sep 03 '16 at 13:29
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Welcome to the club of the victims of mysterious and unexplained downvotes ! – Claude Leibovici Sep 03 '16 at 13:40

operatorerror · Answer 3 · 2016-09-03T13:33:19.877

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Let's compare with a series you probably know converges $\sum 1/n^2$. $$ L=\lim_{n\rightarrow \infty}\frac{2^{-\sqrt{n}}}{1/n^2}= \lim_{n\rightarrow \infty}\frac{n^2}{2^{\sqrt{n}}} $$ Since this is hard to look at, take $t=\sqrt{n}$, then $$ \lim_{t\rightarrow \infty}\frac{t^4}{2^{t}}= 0 $$ By l'hospital's a few times, or just knowing that exponential growth beats polynomial growth for any degree polynomial. So your series converges by comparison to $\sum 1/n^2$

edited Sep 03 '16 at 13:33

answered Sep 03 '16 at 13:22

operatorerror

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Okay thanks, this makes sense. – fejz1234 Sep 03 '16 at 13:46
@fejz1234 No problem! – operatorerror Sep 03 '16 at 13:47
Im trying to understand the solution with the comparision of the integral (the other one). And im a bit confused. The third equality doesn't makes sense to me. Maybe you could help me out? – fejz1234 Sep 03 '16 at 13:48
sure. Which inequality are you referring to? the third in the body of the text? – operatorerror Sep 03 '16 at 13:51
Obs. I mean the second equality. When he goes from the integral of $$ e^{-\sqrt{x}\cdot{ln2}} $$ to where he breaks out 2(ln2)^2 and changes whats inside the integral. – fejz1234 Sep 03 '16 at 13:53
@fejz1234 he performed the substitution in brackets – operatorerror Sep 03 '16 at 13:57
Yeah, but even if i replace t with the original substitution .. I cant seem to follow how that equality is true. – fejz1234 Sep 03 '16 at 13:59
This is what I mean : $$ \int^\infty_0 e^{-\sqrt{x}\log 2} dx $$ $$= 2 (\log 2)^2\int^\infty_0 \sqrt{x}\cdot{log2}\cdot{e^{-\sqrt{x}\cdot{log2}}} dx$$
I have problems seeing how this is true
– fejz1234 Sep 03 '16 at 14:03
Let us continue this discussion in chat. – operatorerror Sep 03 '16 at 14:03

score 2 · Answer 4 · answered Sep 03 '16 at 13:27

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Since @User8128 gave the solution using the integral test (probably the simplest for this problem), let me give another way.

For the case where the ratio test is inconclusive as here $$u_k=2^{-\sqrt{k}}\implies \frac {u_{k+1}}{u_k}=2^{\sqrt{k}-\sqrt{k+1}}=1-\frac{\log (2)}{2} \frac 1 {\sqrt{k}} +O\left(\frac{1}{k}\right)$$ Raabe's test is very convenient.

Considering $$R=k \left(\frac {u_k}{u_{k+1}} -1\right)=k\left(2^{\sqrt{k+1}-\sqrt{k}}-1\right)$$ and expanding again for large values of $k$, you will find $$R=\frac{\log ^2(2)}{8}+\frac{\log (2)}{2} \sqrt{k} +O\left(\frac{1}{k^{1/2}}\right)$$ So $R>1$ then the series will be absolutely convergent.

answered Sep 03 '16 at 13:27

Claude Leibovici

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Why the downvotes? – zarathustra Sep 03 '16 at 13:40
@zarathustra seconded. There are two! What could possibly be wrong with this solution. My only hunch is that it is perceived as too complex, but so what, it is exposing OP (and me) to tools they may not know – operatorerror Sep 03 '16 at 13:45
@qbert. I am surprised by the fact that Raabe's test is not often used. I wonder if it is teached or not. May I ask for your opinion ? – Claude Leibovici Sep 03 '16 at 14:01

score -1 · Answer 5 · answered Sep 03 '16 at 13:02

-1

Hint:

$$\lim\limits_{k \to \infty} 2^{\sqrt{k} -\sqrt{k+1}} = 2^{\lim\limits_{k \to \infty} \sqrt{k} -\sqrt{k+1}}$$

Can you finish from here?

answered Sep 03 '16 at 13:02

Piotr Benedysiuk

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might be worth mentioning why this is true – operatorerror Sep 03 '16 at 13:04
Ah okay, so the limit will tend to 1. But then the ratio test doesn't provide with any information right? Which test could I use else to determine wheter it converges or not? Edit: Yeah I would like to know why its true :P – fejz1234 Sep 03 '16 at 13:04
it is true by continuity of exponentiation. Remember that limits and continuous functions commute. I actually don't think this is an answer to your question – operatorerror Sep 03 '16 at 13:16
It's not an answer to the main question no. – Piotr Benedysiuk Sep 03 '16 at 13:17
how was OP supposed to finish from here? – operatorerror Sep 03 '16 at 13:23
He wasn't able to solve the limit. – Piotr Benedysiuk Sep 03 '16 at 13:25

Convergence of $\sum2^{-\sqrt{k}} $

5 Answers5