determine the convergence:

$\displaystyle

\sum\limits_{n = 1}^\infty {\frac{{\sin (nx)e^{nx} }}{n}} \;\quad on\quad (0,\pi )

$

thanks

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- Aug 16th 2008, 06:34 AMkarimathseries
determine the convergence:

$\displaystyle

\sum\limits_{n = 1}^\infty {\frac{{\sin (nx)e^{nx} }}{n}} \;\quad on\quad (0,\pi )

$

thanks - Aug 16th 2008, 09:38 AMCaptainBlack
This does not converhe for any $\displaystyle x \in (0,\pi)$ . The proof hinges on showing that $\displaystyle \sin(nx) \not\to 0$ as $\displaystyle n \to \infty $. Then the limit of the terms does not go to zero, and so the series cannot converge.

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

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

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

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

Now:

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

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

**UNDER CONSTRUCTION**