Prove that $\displaystyle \Sigma_{n=1}^\infty \frac{x^2}{(x^2+1)^n}$ uniformly convergence at $\displaystyle [a,\infty)$ for $\displaystyle a>0$, and not uniformly convergence at $\displaystyle [-a,a]$ for $\displaystyle a>0$.
Thank you so much!
Prove that $\displaystyle \Sigma_{n=1}^\infty \frac{x^2}{(x^2+1)^n}$ uniformly convergence at $\displaystyle [a,\infty)$ for $\displaystyle a>0$, and not uniformly convergence at $\displaystyle [-a,a]$ for $\displaystyle a>0$.
Thank you so much!
To prove uniform convergence on $\displaystyle [a, \infty)$ you could use the Weierstrass M-test - Wikipedia, the free encyclopedia.
You can use the estimation
$\displaystyle \displaystyle\frac{x^2}{(x^2+1)^n}\leq \frac{x^2}{{n\choose 2} x^4}=\frac{1}{x^2} \frac{1}{{n\choose 2}}\leq \frac{1}{a^2} \frac{1}{{n\choose 2}} $ for n>1
Hope that helps.