I have function $$f(x) = \sum^{\infty}_{n=1} x^n(1-x^n)$$ So I have to prove, that this function is define on $[0, 1]$ and to simplify notation. Also I have to prove is this function uniform convergence on $[0, 1]$ I know for $x=1$ function is 0, but I do not how to write the function for $[0, 1]$


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  • 2
    $\begingroup$ You probably meant to say that $\sum\limits_{n=1}^N x^n (1-x^n)$ converges uniformly on $[0,1]$ as $N\rightarrow \infty$, right? $\endgroup$ – Keen-ameteur Aug 14 at 9:22
  • $\begingroup$ An aside: The series $\sum_{n=1}^\infty x^n(1-x)^n$ does converge uniformly on $[0,1]$.. $\endgroup$ – GEdgar Aug 14 at 12:24

For $x\in(0,1)$, $x^n(1-x^n)=x^n-x^{2n}$ so $$f(x)=\sum_{n=1}^\infty x^n-\sum_{n=1}^\infty x^{2n}=\frac{x}{1-x}-\frac{x^2}{1-x^2}=\frac{x}{1-x^2}.$$ If the convergence is uniform, $f(x)$ must be continuous since $x^n(1-x^n)$ is continuous on $[0,1]$ for all $n\geq 1$, but $$\lim_{x\to1}f(x)=\infty\neq0=f(1),$$ so it is not convergent uniformly.


A series cannot converge uniformly unless the general term tends to $0$ uniformly. Take $x=(1-\frac 1 n)$ to see that the general term does not tend to $0$ uniformly in this case.

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    $\begingroup$ Could you direct me to a proof of this '$f_n(x_n) \not\to 0$ implies not uniform convergence' theorem? $\endgroup$ – Botond Aug 14 at 9:46
  • $\begingroup$ It is obviuous from definition: Let $s_n(x)$ be the partial sums sequence for $\sum f_n(x)$. Then $f_n(x)=s_n(x)-s_{n-1}(x)$ so if $s_n \to s$ uniformly then $f_n \to s-s=0$ uniformly. $\endgroup$ – Kavi Rama Murthy Aug 14 at 9:49
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    $\begingroup$ Thank you, but I was talking about the other one; I guess it sounds like this: let $(f_n)$ be a sequence of functions with pointwise limit $f$ and let $(x_n)$ be a sequence in their common domain with limit $x$. If $f_n(x_n) \not\to f(x)$, then $(f_n)$ is not uniformly convergent. (You used something like this yesterday, with $f_n(x)=x^n(1-x^n)$ and $f=0$.) $\endgroup$ – Botond Aug 14 at 10:01
  • $\begingroup$ This is again from definition; prove by contradiction. If $f_n \to f$ uniformly the n $|f_n(x)-f(x)| <\epsilon$ for all $x$ as long as $n$exceeds some integer $n_0$. This integer does not depend on $x$. We can replace $x$ by $x_n$ to get a contradiction. $\endgroup$ – Kavi Rama Murthy Aug 14 at 10:04
  • $\begingroup$ I see now, if $f_n(x_n)\to A \neq f(x)$ then we have that $||f_n-f||$ does not go to zero. $\endgroup$ – Botond Aug 14 at 10:04

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