I have a sequence of functions:
$$S_n(x) = x\frac{1-x^n}{1-x}, \text{ where } x \in (-1, 1)$$
It is known that:
$$\lim_{n \to \infty}S_n(x) = \frac{x}{1-x}$$
thus $S_n(x)$ converges pointwise. I am to prove that is does not converge uniformly. I tried to estimate:
$$\left|S_n(x) - \frac{x}{1-x}\right|$$ using an $x \in (-1, 1)$ but it didn't work.
I would appreciate any help.


5 Answers 5


The convergence is not uniform on $(-1,1)$, but it is uniform on closed subsets thereof. We first examine the convergence on the closed subinterval $[-r,r]$ for $0<r<1$. After that, we show that the convergence fails to be uniform on $(-1,1)$

Let $\epsilon>0$ be given. Then, we have for $x\in [-r,r]$ for $0<r<1$

$$\begin{align} \left|S_n(x)-\frac{x}{1-x}\right|&=\frac{|x|^{n+1}}{|1-x|}\\\\ &\le \frac{r^{n+1}}{1-r}\\\\ &<\epsilon \end{align}$$

whenever $n>\frac{\log(1-r)+\log(\epsilon)}{\log(r)}-1$

Take $\epsilon=1/2$. Then, for all $N$ we take $x=(1/2)^{1/(n+1)}$ for $0<r<1$ and any $n>N$. Then, we find that

$$\begin{align} \left|S_n(x)-\frac{x}{1-x}\right|&=\frac{|x|^{n+1}}{|1-x|}\\\\ &= \frac{1/2}{1-(1/2)^{1/n+1}}\\\\ &\ge 1/2\\\\ &=\epsilon \end{align}$$

which negates the uniform convergence on the open interval $(-1,1)$

  • $\begingroup$ Nice (and complete), +1. $\endgroup$
    – Clement C.
    Commented Apr 6, 2017 at 17:38
  • $\begingroup$ @ClementC. Thank you CC! $\endgroup$
    – Mark Viola
    Commented Apr 6, 2017 at 17:41
  • $\begingroup$ @Dr.MV Thank you! Very clear and nice solution :) $\endgroup$
    – Hendrra
    Commented Apr 6, 2017 at 19:38
  • $\begingroup$ You're welcome. It's my pleasure. And very pleased to hear that this was useful. -Mark $\endgroup$
    – Mark Viola
    Commented Apr 6, 2017 at 19:50

I think you are on the right way, note that:

$$ \left|S_n(x)- \frac{x}{1-x} \right|=\left| -\frac{x^{n+1}}{1-x} \right| \ge | x^{n+1} | > 0 $$ for every $x \in (0,1) $ and every $n \in \mathbb{N}$. Taking least upper bounds basically gives the solution.


So, you have $$ \left\lvert S_n(x) - S(x)\right\rvert = \left\lvert \frac{1-x^n}{1-x} - \frac{1}{1-x}\right\rvert= \left\lvert \frac{x^n}{1-x}\right\rvert $$ for all $x\in(-1,1)$.

Now, consider $x_n \stackrel{\rm def}{=} 1-\frac{1}{n}\in(-1,1)$. We have $$ \sup_{x\in (-1,1)}\left\lvert S_n(x) - S(x)\right\rvert \geq \left\lvert S_n(x_n) - S(x_n)\right\rvert =\left\lvert \frac{x_n^n}{1-x_n}\right\rvert = \frac{(1-\frac{1}{n})^n}{\frac{1}{n}} = n \left(1-\frac{1}{n}\right)^n \xrightarrow[n\to\infty]{}\infty $$ so $\sup_{x\in (-1,1)}\left\lvert S_n(x) - S(x)\right\rvert\not\xrightarrow[n\to\infty]{}0$, meaning there is no uniform convergence on $(-1,1)$.


Note: If a sequence of functions is bounded by a sequence $M_n$, and it converges uniformly to a function $f$, then $f$ is bounded.

Suppose then that $$ S_n(x)=\frac{x(1-x^n)}{1-x}\stackrel{\text{uniformly}}{\rightarrow} \frac{x}{1-x} $$ Then, as $S_n(x)$ is a finite sum, it is bounded for any given $n$.

However, we have that $\frac{x}{1-x}$ is unbounded on $(-1,1)$, which implies the convergence cannot be uniform.

  • $\begingroup$ is unbounded on $(-1,1)$ $\endgroup$
    – reuns
    Commented Apr 7, 2017 at 3:03
  • $\begingroup$ @user1952009 yes oops thank you $\endgroup$ Commented Apr 7, 2017 at 3:33

First, we simplify the expression. Note that: \begin{align} \left|\frac{x(1-x^n)}{1-x} - \frac{x}{1-x}\right| &= \left|\frac{x(1-x^n) -x}{1-x}\right| \\ &=\left|\frac{-x^{n+1}}{1-x}\right| \\ &= \frac{x}{1-x}\cdot x^{n} \end{align}

We want $\left|S_n(x) - \frac{x}{1-x} \right| < \epsilon$ for given $\epsilon$ and $n \ge N$ for some large enough $N$. For this to be the case, we would need an $N$ such that, for all $n\ge N$: \begin{align*} &\frac{x}{1-x}x^n < \epsilon \\ \iff & x^n < \epsilon\frac{1-x}{x} \\ \iff & n > \frac{\ln\left(\epsilon\frac{1-x}{x}\right)}{\ln x} = \frac{\ln \left(\epsilon(1-x)\right) - \ln x}{\ln x} \end{align*}

But, as $x\to 1$ from the left, $\frac{\ln \left(\epsilon(1-x)\right) - \ln x}{\ln x} \to \infty$. So, no such $N$ exists.

Therefore, $S_n(x)$ does not converge uniformly.


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