# Pointwise and uniform convergence of $\sum_{n=0}^\infty x^n(1-x)$

I need to determine whether

$\sum_{n=0}^\infty x^n(1-x)$

converges pointwise and/or uniformly in $\mathbb{R}$

What I tried so far:
$\sum_{n=0}^\infty x^n(1-x)$ converges only if $|x| < 1$ or if $x=1$.

Does that mean that $\sum_{n=0}^\infty x^n(1-x)$ does not converge pointwise in $\mathbb{R}$ and therefore also not uniformly?

• Yes, that's right. It's a good exercise to show convergence is uniform on $[0,1]$ or on $[a, 1]$ where $-1 < a \le 0$. – fredgoodman Feb 3 '18 at 17:27

## 2 Answers

Indeed your series diverges when $x=-1$ (for instance), and therefore, in particular, it is not an uniformly convergent series of functions from $\mathbb R$ into $\mathbb R$.

Obviously it's $$\sum_{n=0}^{\infty}x^n(1-x)=(1-x)\sum_{n=0}^{\infty}x^n$$ but $\sum_{n\geq 0}x^n$ is a well known series which converges for $|x|<1$. For $x=1$ the whole term is zero, hence it converges in $(-1,1]$. The convergence is uniform in $(a,1]$ for each $a\in (-1,1]$. But uniform convergence implies pointwise convergence.