Is $S(x) = \sum_{n=1}^{\infty} x^2 (1-x^2)^{n-1}$ will uniformly convergent on $[-1,1]$ ? True/false .
My attempt : Yes , $S_n$ will uniformly convergent on $[-1,1]$ if i put $x = \frac{1}{\sqrt n}$ then by $M_n$ Test $||f||_{ \infty}= \lim_{n\rightarrow \infty}\frac{1}{n} (\frac{ n-1}{n})^{n-1}= \lim_{\rightarrow \infty} \frac{1}{n} \frac{1}{e}=0$
Is its True ?
Any hints/solution will be appreciated
thanks u
Note that
$$\sup_{x \in [-1,1]}\left|\sum_{k=n+1}^\infty x^2(1-x^2)^{k-1}\right| \geqslant \sup_{x \in [-1,1]}\sum_{k=n+1}^{2n} x^2(1-x^2)^{k-1} \geqslant \sup_{x \in [-1,1]}nx^2(1-x^2)^{2n} $$
Take $x^2 = 1/(2n)$, that is $x = \pm1/\sqrt{2n} \in [-1,1]$, and it follows that
$$\sup_{x \in [-1,1]}\left|\sum_{k=n+1}^\infty x^2(1-x^2)^{k-1}\right| \geqslant \frac{1}{2}\left(1 - \frac{1}{2n}\right)^{2n}$$
Since the limit of the RHS as $n \to \infty$ is $e^{-1}/2 \neq 0$ the convergence is not uniform on $[-1,1]$.