Infinite series and uniform convergence

Question

I wanted to determine if

$$\sum_{n=1}^{\infty} (-1)^n x^n(1-x) \text{ uniformly converges on } [0,1]$$

I know that if $$ S_m(x) = \sum_{n=1}^{m}(-1)^n x^n(1-x) = \frac{-x(x-1)((-x)^m-1)}{x+1}$$

$$ S_m\rightarrow \frac{x(x-1)}{x+1} \text{ when } m\rightarrow \infty \text{ for } x\in[0,1] $$

$$ \left\lvert S_m(x) - \frac{x(x-1)}{x+1}\right\rvert = \left\lvert \frac{x(x-1)}{x+1}\right\rvert |x^m|$$

I was trying to show that the partial sum of the series was uniformly convergent but I don't know what to do next.

Thanks in advance!

Answer 1

Note $$ \left\lvert S_m(x) - \frac{x(x-1)}{x+1}\right\rvert = \left\lvert \frac{x(x-1)}{x+1}\right\rvert |x^m|\le(1-x)x^m. $$ Let $$ f_m(x)=(1-x)x^m. $$ Let $f_m'(x)=0$ which has a solution $x_m=\frac{m}{m+1}$. Also $f''(x_m)<0$. So $f_m(x)$ attains the max value at $x=x_m$. So $$ \left\lvert S_m(x) - \frac{x(x-1)}{x+1}\right\rvert = \left\lvert \frac{x(x-1)}{x+1}\right\rvert |x^m|\le(1-x)x^m\le f_m(x_m)=\frac1{m+1}(\frac{m}{m+1})^m\to0 $$ as $m\to\infty$. Thus $S_m(x)$ uniformly converges to $\frac{x(x-1)}{x+1}$.

Answer 2

We have uniform convergence by the Cauchy criterion, since

$$\left|\sum_{k=n}^m (-1)^kx^k(1-x)\right| = (1-x)(x^n - (x^{n+1} - x^{n+2}) - \ldots) \leqslant (1 - x)x^n,$$

and as $n \to \infty$,

$$\sup_{x \in [0,1]} (1-x)x^n = \frac{1}{(1 +1/n)^n}\frac{1}{n+1} \to 0 $$