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

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?

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.
• @Gitika You can prove that the series is uniformly convergent on $[-a, a]$ for $a \in (0, 1)$ with the M-test. – Botond Jun 5 at 16:01
• @Gitika Well,if $|x|\leqslant a$ then we have that $|x^n|=|x|^n\leqslant a^n$, and $\sum_{n} a^n$ is convergent if $0<a<1$. – Botond Jun 5 at 18:16
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 $$x\to 1^-.$$