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If one sees the simplification done in equation $5.3$ (bottom of page 29) of this paper it seems that a trigonometric identity has been invoked of the kind,

$$\ln(2) + \sum _ {n=1} ^{\infty} \frac{\cos(n\theta)}{n} = - \ln \left\vert \sin\left(\frac{\theta}{2}\right)\right\vert $$

Is the above true and if yes then can someone help me prove it?

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Are you familiar with writing $\cos$ as a sum of complex exponentials? Are you familiar with the Taylor series expansion of $-\log(1-x)$? –  anon Jul 24 '12 at 21:30
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1 Answer 1

up vote 5 down vote accepted

Hint 1: Use that $$ \log(1-z)=-\sum\limits_{n=1}^\infty\frac{z^n}{n} $$

Hint 2: Set $$ z=r e^{i\theta} $$

Hint 3: Take a real part

Hint 4: Take a limit $r\to 1-0$ and use Abel's summation formula

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Just a small remark : in the second step, some work is required to justify convergence (one might use Abel's summation formula for example). –  Joel Cohen Jul 24 '12 at 21:40
    
...finally: $1-\cos\,\theta=2\sin^2\frac{\theta}{2}$ –  J. M. Jul 24 '12 at 21:40
    
@Joel, yes, since $|\exp\,i\theta|=1$, which is exactly the boundary of the convergence region of the series being used... –  J. M. Jul 24 '12 at 21:41
    
MSE users are always on alert, thanks for your remarks! –  Norbert Jul 24 '12 at 21:50
    
For people not used to Norbert's notation in Hint 4: what he wrote is the same as $r\to1^{-}$ (that is, approach from the left). –  J. M. Jul 24 '12 at 21:55
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