http://www.mathhelpforum.com/math-he...1&d=1241547102

Printable View

- May 5th 2009, 10:11 AMsilversandhow to show this series uni. convergent
- May 5th 2009, 01:15 PMredsoxfan325
Since $\displaystyle |u_n(x)|\leq M_n$, then for large enough $\displaystyle n,m$, we can say that $\displaystyle \left|\sum_{i=n}^m u_i(x)\right|\leq\sum_{i=n}^m M_i < \epsilon$ for all $\displaystyle x$.

Letting $\displaystyle m\to\infty$, we can say that $\displaystyle \exists~N$ such that $\displaystyle n>N \implies |u_n(x)-u(x)|<\epsilon$, so $\displaystyle u_i$ converges uniformly to $\displaystyle u$. - May 9th 2009, 05:50 PMhkito
Weierstrass M-test.