1. uniform convergence

Show that the series ${\Sigma_{n=1}^{\infty}} \frac {sin nx}{n^2}$ uniformly converges on $\mathbb{R}$. Verify that

$\int_{0}^{\pi} ({\Sigma_{n=1}^{\infty}}\frac {sin nx}{n^2}) dx$ = ${\Sigma_{n=1}^{\infty}}\frac {2}{(2n-1)^3}$

2. Because is...

$\displaystyle \int_{0}^{\pi} \sin nx\ dx = \frac{1-\cos n \pi}{n}$ (1)

... if the series $\displaystyle \sum_{n=1}^{\infty} \frac{\sin nx}{n^{2}}$ converges uniformely, then is...

$\displaystyle \int_{0}^{\pi}\ \sum_{n=1}^{\infty} \frac{\sin nx}{n^{2}}\ dx = \sum_{n=1}^{\infty}\ \int_{0}^{\pi} \frac{\sin nx}{n^{2}}\ dx = 2\ \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{3}}$ (2)

At this point it remains to demonstrate the first part of Your statement...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
At this point it remains to demonstrate the first part of Your statement...
Because in all $\mathcal {R}$ is...

$\displaystyle \frac{|\sin nx|}{n^{2}} \le \frac{1}{n^{2}}$ (1)

... and the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}$ converges absolutely, the the series...

$\displaystyle f(x) = \sum_{n=1}^{\infty} \frac{\sin nx}{n^{2}}$ (2)

... converges uniformely in all $\mathcal{R}$...

Kind regards

$\chi$ $\sigma$