So as the title says, can anyone prove that $\{ \sum_{n=0}^k z^n \}_{k=0}^\infty$ does not converge uniformly 0n the disk $D(0,1)$? I think it would converge uniformly to $1/(1-z)$ since it is a geometric series, but professor posed the problem so I'm thinking that must not be correct. Thoughts?

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    $\begingroup$ It converges uniformly on compact subsets of $D(0, 1)$. However, since there's a pole at $z=1$, it can't converges uniformly. $\endgroup$ – Jacky Chong Sep 27 '16 at 3:58
  • $\begingroup$ @Masacroso I don't know what the M-test is. I guess the real root of my question is why doesn't it converge to $1/(1-z)$? $\endgroup$ – Birdman2246 Sep 27 '16 at 4:05
  • $\begingroup$ @JackyChong That's the open disk. I don't know if that would make a difference since we haven't learned about poles yet, but I'm guessing it would. $\endgroup$ – Birdman2246 Sep 27 '16 at 4:06
  • $\begingroup$ I know it's the open disk. My comment is that it can't possibly converge uniformly on the disk, but for any smaller compact set the series will converge uniformly. $\endgroup$ – Jacky Chong Sep 27 '16 at 4:08
  • $\begingroup$ @JackyChong Oh, I see. Since Uniform convergence is checked for every $z \in D(0,1)$, we can get a $z$ very close to $1$ that causes $||f_n - f||$ to become unbounded. Sorry for the misunderstanding; it looks like I just need to work more on understanding Uniform Convergence vs Pointwise Convergence. $\endgroup$ – Birdman2246 Sep 27 '16 at 4:33

Consider $f_n(z) = 1+z+\cdots + z^n = { 1-z^{n+1} \over 1-z}$. Then, for $|z|<1$, we have $f_n(z) \to f(z)={1 \over 1-z}$.

We have $d_n(z)=f(z)-f_n(z) = {z^n \over 1-z}$ (this is pointwise convergence).

Let $z_n = {n \over 1+n}$, and note that $|z_n| <1$ and $d_n(z_n) = {1+n \over (1-{1 \over 1+n})} (1-{1 \over n+1})^{n+1}$. Since $(1-{1 \over n+1})^{n+1} \to {1 \over e}$, we see that $d_n(z_n) \to \infty$.

In particular, $\lim_n \sup_{|z|<1} |f(z)-f_n(z)| = \infty$

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    $\begingroup$ This is a great answer, because it not only shows how to possibly solve it but provides a specific sequence in $D(0,1)$ that causes it to be unbounded. I can't thank you enough. $\endgroup$ – Birdman2246 Sep 27 '16 at 14:46
  • $\begingroup$ @Birdman2246: Glad to be able to help, thanks for the uplifting comment! $\endgroup$ – copper.hat Sep 27 '16 at 15:39


In $D(0,1),$

$$\left|\sum_{n=0}^k z^n - (1-z)^{-1}\right| = \left|\frac{-z^{k+1}}{1-z}\right| = \frac{|z|^{k+1}}{|1-z|} \geqslant \frac{|z|^{k+1}}{1 + |z|} \geqslant \frac{|z|^{k+1}}{2}.$$

Now consider the supremum over $D(0,1)$ in the limit as $k \to \infty.$


Realize that $\sum_{k=0}^{\infty} x = 1/(1-x)$ is unbounded at $x=1$ on $(0,1)$. Also realize that $\sum_{k=0}^n x$ is bounded by $n$ on $(0,1)$. From this, you can conclude that convergence cannot be uniform (do you see how?).

  • $\begingroup$ I think this is the simplest answer by far, but it needs to be rewritten. $\endgroup$ – zhw. Sep 27 '16 at 22:03
  • $\begingroup$ Well, I think it gets closest to the heart of what is going on without needless specifics. Any unbounded function must have unbounded terms converging to it in order for the convergence to be uniform. What do you recommend being rewritten? $\endgroup$ – abnry Sep 28 '16 at 16:52

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