Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ($1$)) $\frac{1}{n}\sum_{j=1}^{n}\sin(jx)= \dots$

share|cite|improve this question
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 $\sum_{j=1}^{\infty}sin(jx)$ – alvoutila Oct 14 '12 at 21:21
up vote 2 down vote accepted


$$S_n(x) = \sum_{k=1}^n \sin(kx) = \dfrac{\cos(x/2) - \cos((n+2)x/2)}{2\sin(x/2)}$$ 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 $0$.

share|cite|improve this answer
But how would you use Cesàro method i.e. $\frac{1}{n} \sum_{j=1}^{n} sin(jx) \rightarrow a$ as $j \rightarrow \infty$, where $a \in N$ with that proof? What is telescopic cancellation? I know that it is A-B+(B-C)+(C-D)=A-D, but hard time in figuring out how to implement it in this case? – alvoutila Oct 14 '12 at 20:12
@alvoutila What you want to show is the following. If $$s_n = \sum_{k=1}^{n} a_k$$ then you want to look at $$S_N = \dfrac{\displaystyle \sum_{n=1}^{N} s_n}N$$ and show that the limit of $S_N$ exists. (This is the Cesaro way.) I have added details to the post above and hope now it is clear. – user17762 Oct 14 '12 at 20:32
If it suites you I come back to comment because I have to first set problem which I try to solve and then compare your ideas with what I have achieved. And maybe then I can formulate some questions for you to answer. I understood your comment above, because I know what is Cesàro way as presented in lectures. But Hint and Edit part are not yet fully understood. – alvoutila Oct 14 '12 at 21:25
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 $s_n=sin(nx)$? – alvoutila Oct 14 '12 at 21:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.