Uniform convergence of $\sum\limits_{n=1}^{\infty} (-1)^n (1-x)x^n$

Question

I need to check the uniform convergence of

$\sum\limits_{n=1}^{\infty} (-1)^n (1-x)x^n$ , $x \in[0,1]$.

I tried to use the Weierstrass criterion, but ended up at

$S_n = \sup|(-1)^n (1-x)x^n| = \frac{n^n}{(n+1)^{(n+1)}} $

and

$\sum\limits_{n=1}^{\infty} S_n$ seems to be divergent, which gives me nothing.

At school we always used only the Weierstrass criterion to prove uniform convergence (unless we wanted to prove that the series is not uniform convergent), but according to the book this series should be uniform convergent, so I suppose that there has to be any way how to prove it by using the Weierstrass criterion.

Thanks for any help or hint.

Answer 1

Assuming that your series is$$\sum_{n=0}^\infty(-1)^n(1-x)x^n\quad\text{($x\in[0,1]$)},$$the convergence is uniform by Dirichlet's test:

• the sequence of partial sums of $\sum_{n=0}^\infty(-1)^n$ is bounded;
• the sequence $\bigl((1-x)x^n\bigr)_{n\in\Bbb N}$ is monotonic, for each $x\in[0,1]$;
• the sequence of functions $\bigl((1-x)x^n\bigr)_{n\in\Bbb N}$ converges uniformly to the null function on $[0,1]$.

Answer 2

$$ S_n(x)=(1-x)\sum^n_{k=0}(-x)^k=(1-x)\frac{1-(-x)^{n+1}}{1+x} $$

For $0\leq x\leq 1$ $$ \Big|S_n(x)-\frac{1-x}{1+x}\Big|=\frac{1-x}{1+x}x^{n+1}\leq (1-x)xx^{n}\leq\big(1-\frac{n}{n+1}\Big)\Big(1+\frac{1}{n}\Big)^{-n} $$

can you finish from here?

Answer 3

For each $0\leq x \leq 1$ and $N\geq 1$, $$ \frac{{1 - x}}{{1 + x}} = \sum\limits_{n = 0}^{N - 1} {( - 1)^n (1 - x)x^n } + ( - 1)^N \frac{{1 - x}}{{1 + x}}x^N . $$ Here $$ \left| {( - 1)^N \frac{{1 - x}}{{1 + x}}x^N } \right| \le (1 - x)x^N \le \left( {1 - \frac{N}{{N + 1}}} \right)\left( {\frac{N}{{N + 1}}} \right)^N \le \frac{1}{2}\frac{1}{{N + 1}}. $$ Thus, your series converges to $x \mapsto \frac{{1 - x}}{{1 + x}}$ uniformly on $0\leq x \leq 1$.