Uniform convergence

[steve.b]

Decide if they uniformly convergent:

$\displaystyle \sum_{n=1}^{\infty}\frac{x^2}{1+n^2x^2}$ on [0,1]
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}e^{-nx^2}$ on $\displaystyle \mathbb R$

Thank you in advance!

 

[galactus]

Decide if they uniformly convergent:

Have you tried the Weiserstrass M-Test?. Abel's Convergence Test?. Dirichlet's Test?.

$\displaystyle \sum_{n=1}^{\infty}\frac{x^2}{1+n^2x^2}$ on [0,1]

This series is convergent and has the form of

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{2}+a^{2}}=\frac{1}{2a}coth(\pi a)-\frac{1}{2a^{2}}$.

In this case,

$\displaystyle a=\frac{1}{x}$ and we get convergence to:

$\displaystyle \frac{\pi x}{2}coth(\frac{\pi}{x})-\frac{x^{2}}{2}$.

i.e. if x=1, then it converges to $\displaystyle \sum_{n=1}^{\infty}\frac{1}{1+n^{2}}=\frac{\pi}{2}coth(\pi)-\frac{1}{2}$

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}e^{-nx^2}$ on $\displaystyle \mathbb R$

This series is also convergent and is directly related to what is known as the dilogarithm.

$\displaystyle Li_{2}(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{2}}$

In this case, $\displaystyle z=e^{-x^{2}}$ and we get:

$\displaystyle Li_{2}(e^{-x^{2}})$.

This is a more advanced topic and can be googled if you're interested.