Show uniform convergence of $\sum\limits_n(1-x^2)^2 x^n$ on $[0,1]$

Question

I'm trying to show that if $u_n(x) = (1-x^2)^2 x^n$ then $\sum u_n$ is uniformly convergent on $[0,1]$.

The only thing I can think to use is Weierstraß M-test to show that $u_n < M_n$ where $\sum M_n$ converges, but I can't find a way to make the $M_n$ independent of $x$.

Answer 1

HINT: We can see the limit $u(x)=(1-x^2)^2\frac1{1-x}=(1+x)^2(1-x)$ from the geometric series. Use that the limit is continuous and all $u_n$ are $\ge 0$.

Answer 2

By studying the derivative of $u_n(x)$, we can find the maximum value on $[0,1]$. We have: $$ u'_n(x) = x^{n-1}(x^2 - 1)\left((n + 4)x^2 - n\right) $$

Thus, $u_n(x)$ reaches its maximum on $[0, 1]$ at: $$ m_n = \sqrt{\frac{n}{n+4}} $$

And we have: $$ u_n(m_n) = \left(\frac{n}{n+4}\right)^{n/2}\frac{16}{(n+4)^2} \le \frac{16}{n^2} = M_n $$

Since $\sum 1/n^2$ converges, it now follows that the convergence is uniform.