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

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.

  • $\begingroup$ 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
  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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.