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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?

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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.

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  • $\begingroup$ What are you saying about the M-test? $\endgroup$ – Gitika Jun 5 at 16:00
  • $\begingroup$ @Gitika You can prove that the series is uniformly convergent on $[-a, a]$ for $a \in (0, 1)$ with the M-test. $\endgroup$ – Botond Jun 5 at 16:01
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    $\begingroup$ You can use cauchy hadamard or the ratio test if you dont like the M test. $\endgroup$ – eminem Jun 5 at 16:13
  • $\begingroup$ @Botond..with which series should I compare to apply the M- test? $\endgroup$ – Gitika Jun 5 at 17:54
  • $\begingroup$ @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 $0<a<1$. $\endgroup$ – Botond Jun 5 at 18:16
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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 $x\to 1^-.$

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