-2
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$$\sum_{n=1}^\infty x(1-x)^{n-1}$$

I know that this sum converge $\iff$ $0\le x \le 1$, i wanted to use the Weierstrass but could not suceed, so i think this sum might not converge uniformly,but i'm having problem showing it.

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  • $\begingroup$ Compute the partial sums explicitly. It is a geometric sum. $\endgroup$ – Kabo Murphy Sep 21 '18 at 9:36
  • $\begingroup$ Note the following two facts: (1) the summands are continuous functions, (2) the limit is not continuous on $[0, 1]$. (Hint. you can compute the limit explicitly. See Kavi Rama Murthy's comment.). What can you say from these? $\endgroup$ – Sangchul Lee Sep 21 '18 at 9:39
  • $\begingroup$ Maybe not relevant, but this one cannot be tested by Weierstrass M test. $\endgroup$ – xbh Sep 21 '18 at 9:43
  • $\begingroup$ @xbh How did you manage to infer that? $\endgroup$ – Maor Rocky Sep 21 '18 at 9:44
  • $\begingroup$ There is an exercise to show that there are some uniformly convergent series cannot be tested by Weierstrass M test. Seems like this one. Maybe i have remembered wrong. If not correct, i will scrap this comment. $\endgroup$ – xbh Sep 21 '18 at 9:47
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  1. The series $\sum_{n=1}^\infty x(1-x)^{n-1}$ converges for $0 \le x <2$ !!

  2. Show that $\sum_{n=1}^\infty x(1-x)^{n-1}=1$ for all $x \in (0,2)$.

  3. Let $s_N(x):=\sum_{n=1}^N x(1-x)^{n-1}$ and show that $|s_N(x)-1|=|1-x|^N$.

  4. Let $x_N:=1-\frac{1}{2^{1/N}}$ and show that $|s_N(x_N)-1| =1/2$ for all $N$.

  5. Conclude from 4. that the series does not converge uniformly.

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Let $f(x)=\sum_{n=1}^\infty x(1-x)^{n-1}$ for $x \in [0,2)$. Then

$f(x)=1$ if $x \in (0,2)$ and $f(0)=0$. Hence $f$ is not continuous on $[0,2)$. Therefore the series does not converge uniformly.

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