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Could anyone give a hint how to prove the convergence of the following sum?

$$\sum_n^\infty (-1)^n\frac{\sin^2 n}n$$

I tried writing it like this instead:

$$\sum_n^\infty \frac1n (-1)^n \sin^2 n.$$

From here, it is easy to see that $\frac1n$ is a bounded and strictly decreasing sequence. It would be sufficient to prove that the sequence of partial sums of $(-1)^n\sin^2 n$ is bounded.

From here, I get that $(-1)^n\sin^2 n = (-1)^n\frac{1 - \cos 2n}2 = \frac{(-1)^n}2 - \frac{(-1)^n \cos2n}2$, where the sequence of partial sums of $\frac{(-1)^n}2$ is bounded as well as the sequence of partial sums of $\frac{\cos 2n}2$. Unfortunately, I cannot tell anything about $\frac{(-1)^n\cos 2n}2$.

Thank you.

Martin Sleziak
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David
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4 Answers4

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We will use Dirichlet's test: http://en.wikipedia.org/wiki/Dirichlet%27s_test

Write $\frac 1 n (-1)^n\sin(n)^2 = \frac 1 n (-1)^{n+1} \frac{e^{2in}-2+e^{-2in}}{4}$, which follows from Euler's formulas.

Our series naturally splits in a sum of $3$ series. Set $a_n = \frac 1 n$. Set $b_n =(-1)^{n+1} \frac{e^{2in}}{4}$. Then $\sum_{n=1}^M b_n $ is a geometric sum, with quotient $-e^{2i}$, that can be evaluated as $\frac{e^{2i}}{4} \frac{(-1)^M e^{2iM}-1}{-e^{2i}-1}$. We see that $|\sum_{n=1}^M b_n|<2$ for all $M$. By Dirichlet's test $\sum a_n b_n$ converges. The remaining two series can be handled similarly.

FanOfFourier
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3

The series is not absolutely convergent, so the study of convergence is of interest.

We have $$\sin^2n=1-\cos^2n=1-\frac{1+\cos(2n)}2=\frac 12-\frac{\cos(2n)}2.$$ Since $$\sum_{n=1}^\infty\frac{(-1)^n}n\mbox{ is convergent},$$ we only have to address the convergence of $$\sum_{n=1}^\infty (-1)^n\frac{\cos(2n)}n,$$ which can be done by a summation by parts. Indeed, we define $s_n:=\sum_{k=0}^n(-1)^k$. Then $$\sum_{k=M}^N(-1)^k\frac{\cos(2k)}k=\sum_{n=M}^Ns_n\frac{\cos(2n)}n-\sum_{n=M-1}^{N-1}s_n\frac{\cos(2(n+1))}{n+1}.$$ Since the series $\sum_k\frac 1{k^2}$ is convergent, we actually only have to show that the series $$\sum_{n=1}^\infty s_n\frac{\cos(2n)}{n}\mbox{ and }\sum_{n=1}^\infty s_n\frac{\cos(2(n+1))}{n}$$ are convergent. (Indeed, $\frac{\cos(2n)}n-\frac{\cos(2(n+1))}{n+1}=\frac{\cos(2n)-\cos(2(n+1))}n-\cos(2(n+1))\left(\frac 1n-\frac 1{n+1}\right) $.) Since $s_{2k+1}=0$, it's enough to establish the convergence of $$\sum_{n=1}^{\infty}\frac{\cos(4n)}n\mbox{ and }\sum_{n=1}^{\infty}\frac{\cos(2(2n+1))}n.$$ It can be done by (an other!) summation by parts.

Community
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Davide Giraudo
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  • I set $a_n = \frac1n$ and $b_n = (-1)^n\cos(2n)$, but I do not see how do I match this onto $\sum_{k=m}^n a_k(b_{k+1} - b_k)$. – David Aug 23 '13 at 13:43
  • @DavidČepelík I've added details. – Davide Giraudo Aug 23 '13 at 13:53
  • I have means of proving that the last two sums are convergent, so I could get away with just one summation by parts. But I do get a little lost where you talk about $\sum_k \frac1{k^2}$. Where does it come from? – David Aug 23 '13 at 14:13
  • I've edited. Is it clearer? – Davide Giraudo Aug 23 '13 at 14:29
  • How would you prove that the last two sums converge. That seems like the difficult part – George Lowther Aug 23 '13 at 16:55
  • Can't we use the fact that $|\sum_{k=1}^N\cos(4k)|$ is bounded, the same for $\sin$? – Davide Giraudo Aug 23 '13 at 16:58
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    @DavideGiraudo: ah, yes, that works. You need to know that $\pi$ is irrational for that (essentially, the same thing as using equidistribution) but it gives convergence. – George Lowther Aug 23 '13 at 17:13
  • Actually, I was wrong there, you don't need to know that $\pi$ is rational. Just writing in terms of complex exponentials and using the formula for summing geometric series works fine on its own. – George Lowther Aug 23 '13 at 20:23
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EDIT $$- \log(1+e^{\pm 2i}) =\sum_{n\in\mathbb{N}} (-1)^n \frac{e^{\pm 2i n}}{n}$$ And $$\sin^2 n = -\frac{1}{4} (e^{in} - e^{-in})^2 = - \frac{1}{4} (e^{2in} - 2 + e^{-2in}) = \frac{1}{2} - \frac{1}{4} (e^{2in} + e^{-2in})$$ So with $\Sigma$ for the "target sum": $$\Sigma = \frac{1}{2} \underbrace{\sum_{n\in\mathbb{N}} (-1)^n \frac{1}{n}}_{=-\log(2)} - \frac{1}{4} \underbrace{\sum_{n\in\mathbb{N}} (-1)^n\frac{e^{2in}}{n}}_{= -\log(1+e^{2in})} - \frac{1}{4} \underbrace{\sum_{n\in\mathbb{N}} (-1)^n \frac{e^{-2in}}{n}}_{=-\log(1+e^{-2in})} = \frac{1}{4} (-\log(4) + \log(1+e^{2i}) + \log(1+e^{-2i}))$$

AlexR
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  • only positive series will work – eccstartup Aug 23 '13 at 11:51
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    I do not think you can use comparison test; the terms of the sequence are not positive. – David Aug 23 '13 at 11:52
  • Okay, but I think you can still get the same convergence behaviour somehow... I see that $a_n = \frac{1}{n} \chi_{{2\mathbb{N}}}$ provides a counter-example. I'll look into it. Thanks for the remark. – AlexR Aug 23 '13 at 11:57
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    Approximating the value of the $n$-th term as $n \to \infty$ won't probably work either; you are summing a (very small) error infinitely many times; hence you can get different results. – David Aug 23 '13 at 11:58
  • I removed the false portion to avoid irritations. – AlexR Aug 23 '13 at 12:06
  • I am sorry, what is WIA? – David Aug 23 '13 at 12:08
  • I see - W|A - WolframAlpha. Thought it's Windows Image Acquisition, looked really strange in the context of a math post :). – David Aug 23 '13 at 12:15
  • @DavidČepelík Is that okay? – AlexR Aug 23 '13 at 12:37
  • @AlexR Thanks for your answer. I would like to wait a little longer. Maybe another proof will arise that is simpler. Unfortunately, I haven't had my complex analysis class yet, so my understanding of this proof is somewhat limited. But from what I can comprehend, I think it is correct this way. – David Aug 23 '13 at 13:05
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    Actually, the argument seems a bit circular: you used the expression given in the first equation, but I'm not sure proving the convergence of the series in the RHS is simpler than proving those of the OP. – Davide Giraudo Aug 23 '13 at 14:35
  • @DavideGiraudo I use the Series expansion of the logarithm, yes, but that and eulers' identity for sine is all I use here. I don't see any "circular" proof... – AlexR Aug 23 '13 at 21:23
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    You have to prove that the series expansion of the logarithm is convergent in this part of boundary of the disk of convergence, which is not easier than the proof of the convergence of $\sum_n(-1)^n\frac{\sin n}n$. (what I mean is that it deserves some details) – Davide Giraudo Aug 23 '13 at 21:28
  • Well, from my point of view, the series expansion of the Logarithm is somewhat "basic knowledge". If you don't allow this Identity, then you need to find another proof (for example yours) ;) – AlexR Aug 23 '13 at 21:29
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Using Alternative convergence theorem. As long As $b_n$ has limit when $n$ approaches infinity, and the limit equals to $0$, the serise $a_n=(-1)^n b_n$ is convergence.

J.Johnny 1905
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