checking uniform convergence of series $\sum_{n=1}^\infty x^n$

Question

I have a doubt in a question in which I need to check the uniform convergence of the series given by:

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

Now if the series is uniformly convergent,then its sequence of partial sums (s$_n$) is uniformly convergent.

I have found that $s$$_n$ = $\frac{1-x^n}{1-x}$ which point-wise converges to $s$($x$) = $\frac{1}{1-x}$

If ($s$$_n$) is uniformly convergent ,then $\sup${$\lvert s_n(x)-s(x)\rvert$:$x$ $\in$($-1,1$)| should tend to $0$ as $n$ tends to infinity.

Now how to check whether $\sup${$\lvert $$\frac{x^n}{1-x}\rvert$:$x$ $\in$($-1,1$)} converges to zero as $n$ tends to infinity or how can I use the definition here?

Answer 1

The supremum is $+\infty$ for all $n$, because $$\lim_{x \to 1-0}\frac{x^n}{1-x}=+\infty$$ The best you can do is uniform convergence on $[-a, a]$ for all $a \in (0, 1)$, with the Weierstass M-test, for example.

Answer 2

Each partial sum of the series is bounded. If the series converged uniformly on $(-1,1),$ then the sum would be bounded there. But the sum is $\dfrac{x}{1-x},$ which is unbounded as $x\to 1^-.$