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Show that if $x \neq 0,\pm 2 \pi,\pm 4 \pi, \dots$, then

$\frac{1}{\tan(x/2)}=2 \sum_{j=1}^{\infty}\sin(jx)$

in Cesàro way/sense. Some hint whether to manipulate

$\sum_{j=1}^{\infty}a_j(x)=\sum_{j=1}^{\infty}\sin(jx) \tag1$

into (using partial sum of ($1$)) $\frac{1}{n}\sum_{j=1}^{n}\sin(jx)= \dots$

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    Would someone edit it better if I have lost some thing when I edited it lately. I don't know why it does not put $\sum_{j=1}^{\infty}a_j(x)$ after $\sum_{j=1}^{\infty}sin(jx)$2012-10-14

1 Answers 1

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HINT

$S_n(x) = \sum_{k=1}^n \sin(kx) = \dfrac{\cos(x/2) - \cos((n+2)x/2)}{2\sin(x/2)}$ To prove this multiply, $S_n(x)$ by $\sin(x/2)$ and make use of the fact that $\sin(A) \sin(B) = \dfrac{\cos(A-B) - \cos(A+B)}2$ and do telescopic cancellation.

EDIT $\dfrac{\displaystyle \sum_{n=1}^{N} S_n(x)}N = \dfrac1{2 \tan(x/2)} - \dfrac{\displaystyle \sum_{n=1}^{N} \cos((n+2)x/2)}{2N \sin(x/2)}$

The second term without the $N$ in the denominator is bounded (Why? One way is to evaluate the sum in a similar spirt as above or write it as exponential and use geometric series to see that it is bounded when $x \neq k \pi$). Hence, if you take the limit as $N \to \infty$, the second term will tend to $0$.

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    firstly I think that you don't have to care constant 2 at all. I mean it is all about $\sum_{n=1}^{N}s_n$, where $s_n=sin(nx)$?2012-10-14