# series

• Aug 16th 2008, 06:34 AM
karimath
series
determine the convergence:

$

$

thanks
• Aug 16th 2008, 09:38 AM
CaptainBlack
Quote:

Originally Posted by karimath
determine the convergence:

$

$

thanks

This does not converhe for any $x \in (0,\pi)$ . The proof hinges on showing that $\sin(nx) \not\to 0$ as $n \to \infty$. Then the limit of the terms does not go to zero, and so the series cannot converge.

A proof that $\sin(nx) \not\to 0$ as $n \to \infty$ can be constructed by supposing otherwise, that is we suppose (for fixed $x$) that:

$\sin(nx) \to 0$ as $n \to \infty$.

This means that for all $\varepsilon>0$ there exists an $N_{\varepsilon}$ such that for all $n>N_{\varepsilon}$ :

$|\sin(nx)|<\varepsilon$

Now:

$|\sin((n+1)x)|=|\sin(nx)\cos(x)+\cos(nx)\sin(x)|$

but by making $\varepsilon$ sufficiently small we can make the right hand side of the last equation as close to $|\sin(x)|$ as we like for $n>N_{\varepsilon}$, but this contradicts the consequence of our assumption that $|\sin((n+1)x)|<\varepsilon$ for sufficently small $\varepsilon$

UNDER CONSTRUCTION