# $\sum_{n=1}^\infty x(1-x)^{n-1}$ Does this sum converge uniformly?

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

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.

• Compute the partial sums explicitly. It is a geometric sum. – Kabo Murphy Sep 21 '18 at 9:36
• 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? – Sangchul Lee Sep 21 '18 at 9:39
• Maybe not relevant, but this one cannot be tested by Weierstrass M test. – xbh Sep 21 '18 at 9:43
• @xbh How did you manage to infer that? – Maor Rocky Sep 21 '18 at 9:44
• 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. – xbh Sep 21 '18 at 9:47

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.

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.