steve.b said:
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.
