1

I have two sequences as follows:

$\sum\limits_{n=1}^\infty \frac{\sin n\phi}{n}$
$\sum\limits_{n=1}^\infty \frac{\cos n\phi}{n}$

How to investigate convergence of those two? What criteria should i apply (most likely simmilar one in each of those two cases but still i do not know which)?

kurkowski
  • 639
  • try this: http://math.stackexchange.com/questions/28830/does-sum-frac-sinnxn-converge-uniformly-for-all-x-in-0-2-pi – jh4 Dec 08 '14 at 12:58

1 Answers1

0

For any $\color{blue}{\phi\in(0,2\pi)}$ both series converge due to Dirichlet's criterion:

If $\{a_n\}_{n\in\mathbb{N}}$ is a sequence with bounded partial sums and $\{b_n\}_{n\in\mathbb{N}}$ is a decreasing sequence that converges to zero, then $\sum a_n b_n$ is a converging series.

Moreover, by applying Abel's lemma to the Taylor series of $\log(1-x)$: $$-\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n}$$ we get:

$$\sum_{n\geq 1}\frac{\sin(n\phi)}{n}=-\Im\log\left(1-e^{i\phi}\right)=\frac{\pi-\phi}{2},$$

and:

$$\sum_{n\geq 1}\frac{\cos(n\phi)}{n}=-\Re\log\left(1-e^{i\phi}\right)=-\log\left(2\sin\frac{\phi}{2}\right).$$

Jack D'Aurizio
  • 353,855
  • but what about convergence for $\phi \in \mathbb{R}$? then it is not? I was also thinking about rewriting this (first sum in my question) into $\frac{\sin \frac{n+1}{2} \phi \sin \frac{n\phi}{2}}{n \sin \frac{n\phi}{2}}$ what can i do with this expression? because i think this will be easier after this transformation – kurkowski Dec 08 '14 at 13:45
  • @kurkowski: You can always assume $\phi\in[0,2\pi)$ since the series are unaffected by replacing $\phi$ with $\phi\pm 2\pi$. – Jack D'Aurizio Dec 08 '14 at 13:51
  • And what about my idea of changing the first sum into such an expression? Will it work? – kurkowski Dec 08 '14 at 14:04