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Does the following converge or diverge? $$ \sum_{n=2}^{\infty} a_n^{-n}, $$ where$$ a_n = \int_1^n \sin{\left(\frac{1}{\sqrt{x}}\right)} \,\mathrm{d}x. $$ My friends thought this sum would converge. I think we should do the square root test, checking the value of $t=\lim\limits_{n \to \infty} \dfrac{1}{a_n}$. And we knew that this goes to $0$. So, that would make the sum to converge…

Is there any problem in my ideas? Some said that this sum diverges, which I can never understand why…

It will be great if someone can explain me on this… Better with some proofs I can understand. (We are students learning calculus.)

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  • $\begingroup$ You should find out what happens to $a_n$ as $n$ gets large. Is it increasing in $n$? Does it ever get larger than 2? $\endgroup$ – Michael Mar 23 '18 at 8:36
  • $\begingroup$ What I found out is that a_n goes to infinity when n gets large... $\endgroup$ – Gratus Mar 23 '18 at 8:37
  • $\begingroup$ So you know $a_n \geq 2$ for all sufficiently large $n$. Can you infer anything from that, like a comparison? $\endgroup$ – Michael Mar 23 '18 at 8:38
  • $\begingroup$ Yes I thought that it would converge that way. Isn't it? $\endgroup$ – Gratus Mar 23 '18 at 8:40
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    $\begingroup$ Using that, we know that without finite number of cases a_n^(-n) is smaller than 2^(-n) so we can use the comparison test to prove that the original sum converges... $\endgroup$ – Gratus Mar 23 '18 at 8:42
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Your approach is correct: $a_n$ is positive and by applying the root test the convergence of the series $\sum_n (1/a_n)^{n}$ follows as soon as you show that $1/a_n\to L<1$. Here we have that $1/a_n\to 0$ so $\sum_n (1/a_n)^{n}$ converges.

In fact, $\sin(t)\geq 2t/\pi$ for $t\in [0,\pi/2]$ ($\sin(x)$ is concave in that interval). Hence for $n>1$, $$a_n = \int_1^n \sin{\left(\frac{1}{\sqrt{x}}\right)} \,dx\geq \frac{2}{\pi}\int_1^n \frac{1}{\sqrt{x}}\, dx =\frac{4(\sqrt{n}-1)}{\pi}\implies 0<\frac{1}{a_n}\leq \frac{\pi}{4(\sqrt{n}-1)}$$ which implies that $1/a_n\to 0$ as $n$ goes to infinity.

P.S. By the way $\sin(t)\leq t$ for $t\geq 0$ implies that $$a_n = \int_1^n \sin{\left(\frac{1}{\sqrt{x}}\right)} \,dx\leq \int_1^n \frac{1}{\sqrt{x}}\, dx ={2(\sqrt{n}-1)}\implies \frac{1}{a_n}\geq \frac{1}{2(\sqrt{n}-1)}$$ which implies that $\sum_n (1/a_n)=+\infty$.

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  • $\begingroup$ So, 1/a_n is smaller than that. I understood it... But I don't get how we get the solution by that. Pi/(4(sqrtn-1) is still diverging... $\endgroup$ – Gratus Mar 23 '18 at 8:48
  • $\begingroup$ Thanks. I understood what you meant $\endgroup$ – Gratus Mar 23 '18 at 8:52
  • $\begingroup$ Are you sure that the given series is $\sum_{n=2}^{\infty} a_n^{-n}$? Note that the series $\sum_{n=2}^{\infty} a_n^{-1}$ is divergent. $\endgroup$ – Robert Z Mar 23 '18 at 8:54
  • $\begingroup$ Yeah given series was sigma (a_n)^(-n) $\endgroup$ – Gratus Mar 23 '18 at 9:04
  • $\begingroup$ @GSS Ok. Anyway I left a comment in my P.S. $\endgroup$ – Robert Z Mar 23 '18 at 17:26
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From $\sin x<x$, and for $n>4$,

$$\left(\int_1^n\sin\frac1{\sqrt x}dx\right)^{-n}<\left(2\sqrt n\right)^{-n}<\frac1{n^2}$$

and the series converges.

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  • $\begingroup$ +1 And compliments for your way to reach the $10^5$! Awesome! $\endgroup$ – Von Neumann Mar 23 '18 at 9:20
  • $\begingroup$ @VonNeumann: I have rewritten in a more concise way (you can withdraw your vote if you think it got worse :-). Thanks for the comment ! $\endgroup$ – Yves Daoust Mar 23 '18 at 9:25
  • $\begingroup$ On the contrary, I like the "new" approach very much! $\endgroup$ – Von Neumann Mar 23 '18 at 9:25
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As $n\to +\infty$ you have that asymptotically

$$\int_1^n\sin\frac{1}{\sqrt{x}} \sim 2\sqrt{n}$$

Thence the related series goes like $\frac{1}{n^{n+1/2}}$ which converges.

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  • $\begingroup$ Be aware that he takes the power with -n. It would diverge if he sums up alle the inverses though. $\endgroup$ – MrMatzetoni Mar 23 '18 at 8:46
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    $\begingroup$ Can't understand the downvote nor the above comment. Was the question changed ? $\endgroup$ – Yves Daoust Mar 23 '18 at 9:09
  • $\begingroup$ @YvesDaoust No the question wasn't changed, and I don't understand the down vote (and the previous comment) either... I think it's become a fashion sometimes (the down-voting) $\endgroup$ – Von Neumann Mar 23 '18 at 9:19
  • $\begingroup$ @VonNeumann : I am flagging this answer for moderator, only because it seems there may be a Stackexchange bug (not because you or anyone did anything wrong, and certainly there is no reason to remove your answer). I will explain my concern in my next comment, perhaps you or a moderator could illuminate. $\endgroup$ – Michael Mar 23 '18 at 16:37
  • $\begingroup$ Von Neumann or moderator: Perhaps stackexchange bug since this answer has no edit trail such as "edited 7 hours ago." As I recall, this answer was originally given with an opposite conclusion. I believe the downvote and the MrMatzetoni comment were given then. Next, Von Neumann wrote a message "I am deleting because I misread the question." Then Von Neumann deleted. This morning I observe the answer re-appears, with downvote and comment retained, but no edit trail shown. Stackexchange presents this as if the answer was given originally in its current form, which is not the case. $\endgroup$ – Michael Mar 23 '18 at 17:10

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