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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.

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somewhat related to math.stackexchange.com/questions/273559/… –  Shobhit Aug 23 '13 at 11:35
    
Thanks. I know this proof, but I will have a look at it again. –  David Čepelík Aug 23 '13 at 11:36
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Why not use Leibnitz and $$0 \leq \frac{\sin^2 n}{n} \leq \frac{1}{n}$$ ?? –  AlexR Aug 23 '13 at 11:39
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@AlexR You'd have to have that $\frac{\sin^2 n}n$ is a monotonic sequence, and you do not have that. At least I know Leibniz's theorem like this. –  David Čepelík Aug 23 '13 at 11:44
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Use it on $\frac{1}{n}$ and combine with comparison ;) –  AlexR Aug 23 '13 at 11:47

4 Answers 4

up vote 3 down vote accepted

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.

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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 Čepelík 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 Čepelík Aug 23 '13 at 14:13
    
I've edited. Is it clearer? –  Davide Giraudo Aug 23 '13 at 14:29
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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

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.

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Hold a minute, please; it's $\sin^2 n = (\sin n)^2$, not $\sin n^2$. Or is it a typo? –  David Čepelík Aug 23 '13 at 12:13
    
I got lazy for the $\frac{1}{n}$-Terms, but that suffices for the convergence part =) –  AlexR Aug 23 '13 at 12:19
    
@David, I meant to use $(\sin n)^2$ in my answer. That is what I mean by $\sin(n)^2$. (Note that parentheses.) –  FanOfFourier Aug 23 '13 at 12:40

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.

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Could you please support this statement with a link to a page describing the theorem you are using? –  David Čepelík Aug 23 '13 at 13:01
    
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Unfortunately, you still need to have that $b_n$ is a nonincreasing sequence; and you do not have that as neither $(-1)^n$ nor $\frac{sin^2 n}n$ are monotone. (If you let $b_n := \frac1n$ for $n$ even and zero otherwise, it would not work, for example. The limit would be zero, obviously, but the sum would not converge.) Thanks for help anyway. –  David Čepelík Aug 23 '13 at 13:10
    
see the second one. –  J.Johnny 1905 Aug 23 '13 at 13:12
    
I have seen both. –  David Čepelík Aug 23 '13 at 13:47

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": $$\Sigma = \frac{1}{2} \underbrace{\sum_{n\in\mathbb{N}} (-1)^n \frac{1}{n}}_{=-\log(2)} - \frac{1}{4} \underbrace{\sum_{n\in\mathbb{N}} (-1)^n\frac{e^{2in}}{n}}_{= -\log(1+e^{2in})} - \frac{1}{4} \underbrace{\sum_{n\in\mathbb{N}} (-1)^n \frac{e^{-2in}}{n}}_{=-\log(1+e^{-2in})} = \frac{1}{4} (-\log(4) + \log(1+e^{2i}) + \log(1+e^{-2i}))$$

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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 Čepelík Aug 23 '13 at 11:52
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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 Čepelík Aug 23 '13 at 11:58
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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
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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

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