Uniform Convergence of $\sum_{n=0}^\infty (1-x^2)^2x^n$ on $[0,1]$; subsequent integral

Question

Let $a_n = (1-x^2)^2 x^n$. Show that $\sum_{n=0}^\infty a_n$ converges uniformly on $[0,1]$ and deduce that $\int_0^1 \frac{(1-x^2)^2}{1-x} dx = \sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)}$.

Attempt: Denoting partial sums by $S_n$, we have \begin{eqnarray*} S_n &=& \sum_{k=0}^n (1-x^2)^2 x^k \\ &=&(1-x^2)^2 \frac{1-x^{n+1}}{1-x} \\ &=& (1+x)(1-x^2)(1-x^{n+1}). \end{eqnarray*} I tried to show uniform convergence by showing convergence of $\sup_{x \in [0,1]} S_n(x)$, but I seem unable to solve the resulting first order condition to find the maximizing x. Is there a simpler way to do this?

If I could show uniform convergence, I could integrate the infinite series term by term yielding \begin{eqnarray*} \int \frac{(1-x^2)^2}{1-x} dx &=& \sum_{n=0}^\infty \int_0^1 (1-x^2)^2 x^n dx \\ &=&\sum_{n=0}^\infty \int_0^1 x^n - 2x^{2+n} + x^{4+n} dx \\ &=& \dots \\ &=&\sum_{n=0}^\infty \frac{8}{(n+1)(n+3)(n+5)} \\ &=&\sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)} \end{eqnarray*} and I would be done.

Answer 1

\begin{eqnarray*} S_n &=& \sum_{k=0}^n (1-x^2)^2 x^k \\ &=&(1-x^2)^2 \frac{1-x^{n+1}}{1-x} \\ &=& (1+x)(1-x^2)(1). \end{eqnarray*}

the uniform convergence follows from
\begin{eqnarray*} S_m -S_n&=& \sum_{k=n+1}^m (1-x^2)^2 x^k \\ &=&(1-x^2)^2 \frac{x^{n+1}-x^{m+1}}{1-x}\xrightarrow[n \to \infty]{} 0 \\ \end{eqnarray*} Since $x^{m+1} \leq x^{n+1} \to 0$

Edit: To get uniform convergence one needs further to note that $x^{m+1},x^{n+1} \leq 1$ and that $1-x^2 = 1+x (1-x) $ so $$(1-x^2)^2 \frac{x^{n+1}-x^{m+1}}{1-x} = (1+x)^2(1-x)x^{n+1}-x^{m+1}.$$

So for $x>1-\delta$

$$(1+x)^2(1-x)x^{n+1}-x^{m+1}\leq 2\delta $$

and for $x<1-\delta$ $x^n \leq (1-\delta)^n \to 0$

$$(1+x)^2(1-x)x^{n+1}-x^{m+1} \leq 2 (1-\delta)^n$$

Answer 2

To show uniform convergence of this series we can use the Weierstrass M-test. How big can $f_n(x)=(1-x^2)x^n$ be on $[0,1]?$ Set $f_n'(x) = 0.$ We get $x=\sqrt {n/(4+n)}.$ At this $x$ we have

$$ f_n(\sqrt {n/(4+n)}) = [1-n/(4+n)]^2\cdot[\sqrt {n/(4+n)}\,]^n.$$

The first factor on the right is $[4/(4+n)]^2.$ The second one is $\le1$ for all $n.$ Thus $0\le f_n(x) \le [4/(4+n)]^2, x\in [0,1].$ Since $\sum [4/(4+n)]^2<\infty,\sum f_n$ converges uniformly on $[0,1]$ by Weierstrass M.