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How to prove that $\sum\limits_{n=1}\frac{\sin^2n}n$ does not converge without using the series expansion and with the following tests only or a combination of them (comparison or limit comparison test )

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    $\begingroup$ If Dirichlet's test were allowed, I would say rewrite $\sin^2 n = \frac{1}{2}(1-\cos(2n))$ and then you get the sum of a divergent series and a convergent series. $\endgroup$ May 2, 2019 at 22:58
  • $\begingroup$ @DonThousand but this required complex analysis which is above the question level $\endgroup$
    – asmmo
    May 2, 2019 at 23:11
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    $\begingroup$ @DonThousand I don't see the relevance to this question. $\frac{\sin^2 x}{x}$ isn't a decreasing function so the integral test won't be applicable - and for example, $\int_1^n \frac{\sin^2(\pi x)}{x} dx$ is also asymptotic to $\frac{1}{2}\log(n)$. $\endgroup$ May 2, 2019 at 23:19

2 Answers 2

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There is a sequence of positive integers $\{a_n\}=\{1,2,4,5,8,\ldots\}$ where for each $a_n$ in this sequence, $\sin^2a_n$ is greater than $\frac12$, and this is the maximal such sequence.

How frequent are these $a_n$? In every interval $[(k-1)\pi,k\pi]$, there is a subinterval in the middle of length $\pi/2$ where $\sin^2(x)>\frac12$. Since $\pi/2>1$, there is always an integer in this subinterval. So in the intervals $[0,\pi],[\pi,2\pi],[2\pi,3\pi],\ldots$ you can always find (at least) one $a_n$. Let $b_n$ be some integer in the intersection of $[(n-1)\pi,n\pi]$ with $\{a_n\}$. Note $b_n<n\pi$.

Consider $$ \begin{align} \sum_{n=1}^{\infty}\frac{\sin^2(n)}{n}& >\sum_{n=1}^{\infty}\frac{\sin^2(a_n)}{a_n}&&\text{just summing over fewer positive terms}\\ &>\frac12\sum_{n=1}^{\infty}\frac{1}{a_n}&&\text{property of $a_n$}\\ &\geq\frac12\sum_{n=1}^{\infty}\frac{1}{b_n}&&\text{just summing over fewer positive terms}\\ &>\frac1{2\pi}\sum_{n=1}^{\infty}\frac{1}{n}&&\text{property of $b_n$} \end{align}$$

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    $\begingroup$ I think, there was no need to define the $a_n$. You can just choose $b_n$ to be an integer in the subinterval in the middle. $\endgroup$ May 3, 2019 at 16:41
  • $\begingroup$ @JulianMejia If you are going for a most streamlined argument, yes. It was more natural (to me) to first consider the existence of the $a_n$, and then get a hold on their somewhat unpredictable frequency in a second stage. Sometimes I prefer a more natural argument over a streamlined one. $\endgroup$
    – 2'5 9'2
    May 3, 2019 at 18:48
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Sorry this would be only a comment if it weren't for my low reputation since it is at best a proof sketch, but I believe we may have (in my class) looked at all the natural numbers that are better and better approximations of $2\pi k+\frac{\pi}{2}$, which makes the top strictly greater than the first one in that sequence, and found that subsequence to be divergent since it was a not-too-horrible-to-analyze divergent subsequence of $\sum_{i=1}^{+\infty} \frac{k}{n}$.

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