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I recently saw that the claim that

$\sin(\theta) +\frac{1}{3}\sin(3\theta) +\frac{1}{5}\sin(5\theta)+ \ldots$

converges to $\frac{\pi}{4}$ for $0<\theta<\pi$. Has anyone seen this problem or have any suggestions on how to approach such an interesting problem?

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4 Answers 4

up vote 3 down vote accepted

This is true, (and interesting). From the start, intuition tells me to consider the series for $\arctan(x)$ since it is only over odd integers, and has the correct denominator. This then leads us to $\tanh^{-1}(x)$ to remove to negative signs, and using some identities for this function we can solve the problem.

Solution:

Since $$\tan^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}x^{2n+1}$$we see that $$\frac{1}{i}\tan^{-1}(ix)=\sum_{n=0}^{\infty}\frac{1}{2n+1}x^{2n+1}=\tanh^{-1}(x).$$As the original series is equal to $$\frac{1}{2i}\left(\sum_{n=0}^{\infty}\frac{1}{2n+1}e^{i(2n+1)\theta}-\sum_{n=0}^{\infty}\frac{1}{2n+1}e^{-i(2n+1)\theta}\right)$$we then have $$\sum_{n=0}^{\infty}\frac{\sin\left((2n+1)\theta\right)}{2n+1}=\frac{1}{2i}\left(\tanh^{-1}\left(e^{i\theta}\right)-\tanh^{-1}\left(e^{-i\theta}\right)\right).$$ Now since $$\tanh^{-1}\left(x\right)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$$ we have $$\sum_{n=0}^{\infty}\frac{\sin\left((2n+1)\theta\right)}{2n+1}=\frac{1}{4i}\left(\log\left(\frac{1+e^{i\theta}}{1-e^{i\theta}}\right)-\log\left(\frac{1+e^{-i\theta}}{1-e^{-i\theta}}\right)\right).$$ Hence $$\sum_{n=0}^{\infty}\frac{\sin\left((2n+1)\theta\right)}{2n+1}=\frac{1}{4}\log\left(\frac{e^{i\theta}-e^{-i\theta}}{e^{-i\theta}-e^{i\theta}}\right)=\frac{1}{4i}\log(-1)=\frac{\pi}{4}$$ as desired.

Hope that helps,

Note: The early rearrangements here need justification as we do not have absolute convergence. This is allowable since both series still converge, and the partial sums are identical. Also, the expression for $\tanh^{-1}(x)$ only holds when $|x|<1$, so again justification is required to explain why we can extend to all of $|x|=1$ except the points $x=\pm 1$. This follows from Abel's Limit Theorem.

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Well, this is simply the Fourier series for the odd function of period $2\pi$ which equals $\pi/4$ on the interval $(0,\pi)$.

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Almost same as Eric's answer (using $\log(1+x)$ instead of $\tan^{-1}(x)$. You can argue the convergence of the series to $\log(1+x)$ using generalized alternate series test even though $|e^{inx}| = 1$)

$\sin(n \theta) = \frac{\exp(i n \theta) - \exp(-i n \theta)}{2i}$.

$\displaystyle \sum_{n = \text{odd}} \frac{x^n}{n} = \frac{\log(1+x) - \log(1-x)}{2}$

$\displaystyle \sum_{n=\text{odd}}^{\infty} \frac{\exp(i n \theta) - \exp(-i n \theta)}{2in} = \displaystyle \frac1{2i} \left( \sum_{n=\text{odd}}^{\infty} \frac{\exp(i n \theta)}{n} - \sum_{n=\text{odd}}^{\infty} \frac{\exp(-i n \theta)}{n} \right)$

$= \frac1{4i} \left( \log(1 + \exp(i \theta)) - \log(1 - \exp(i \theta)) - \log(1 + \exp(-i \theta)) + \log(1 - \exp(-i \theta)) \right)$

$= \frac1{4i} \log \left(\frac{(1 + \exp(i \theta))(1 - \exp(- i \theta))}{(1 - \exp(i \theta))(1 + \exp(- i \theta))} \right) = \frac1{4i} \log \left( \frac{\exp(i \theta) - \exp(-i \theta)}{\exp(-i \theta) - \exp(i \theta)} \right) =\frac1{4i} \log(-1) = \frac{\pi i}{4i} = \boxed{\frac{\pi}{4}}$

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1  
$\log(-1)$ should not be evaluated without some extra attention. –  Phira Apr 23 '11 at 20:16
    
Nice way to simplify the early steps! –  Eric Naslund Apr 24 '11 at 0:30

Hint only (in a rush, sorry): Alternatively you might also want to use the Abel summation formula and use cosines.

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1  
Since the question is over half a year old, perhaps you hadn't needed to rush quite that much ... –  Henning Makholm Oct 23 '11 at 6:16

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