Prove uniform convergence of $\sum_{n=1}^{\infty}x^{n}(1-x^{n})$

Question

I need to prove that $\sum_{n=1}^{\infty}x^{n}(1-x^{n})$ converges for $x \in (-1,1]$. Also, I need to prove that $\sum_{n=0}^{\infty}x^{n}(1-x^{n})$ converges uniformly for $x \in (-1+\delta,1-\delta)$ such that $0<\delta<\frac{1}{2}$. Here's my trial:

\begin{align*} \sum_{n=1}^{\infty}x^{n}(1-x^{n}) &= \sum_{n=1}^{\infty}x^{n}- x^{2n}\\ & = (x - x^{2}) + (x^{2} - x^{4}) + (x^{3} - x^{6}) + ...\\ &= x + x^{3} + x^{5} + ...\\ &= \sum_{n=1}^{\infty}x^{2n-1}\\ &= \frac{x}{1-x^2} \end{align*} With the above result, $\sum_{n=0}^{\infty}x^{n}(1-x^{n})$ converges for $x \in (-1,1)$ (but not for $x=1$). I'm open to any ideas for uniform convergence.

Answer 1

For the uniform convergence part, use the Weierstrass M-test:

Suppose that $f_n$ is a sequence of real or complex valued functions defined on a set $A$, and that there is a sequence of non-negative numbers $M_n$ satisfying the conditions $|f_{n}(x)|\leq M_{n}$ for all $n\geq 1$ and all $x\in A$, and $\sum _{n=1}^{\infty } M_{n}$ converges. Then the series $$\sum _{n=1}^{\infty }f_{n}(x)$$ converges absolutely and uniformly on A. (Source: Wikipedia)

To find a sequence $M_n$ in your problem, we have for all $x \in (-1+\delta, 1-\delta )$,

$$ |f_n(x)|=|x^n (1-x^n)| \leq 2\,|x^n| \leq 2\,(1-\delta)^n,$$

and the series $\sum_{n=1}^{\infty} (1-\delta)^n $ converges to $\frac{1}{\delta}$. Therefore by the Weierstrass M-test the series $\sum_{n=1}^{\infty} x^n (1-x^n)$ converges uniformly on that interval.