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Given the series $\sum_{k=0}^\infty z^k $, it is easy to see that it converges locally, but how do I go about showing that it does not also converge uniformly on the open unit disc? I know that for it to converge uniformly on the open disc that $sup{|g(z) - g_k(z)|}$, z element of open unit disc, must equal zero. However, I am finding it difficult to show that this series does not go to zero as k goes to infinity. Edit:Fixed confusing terminology as mentioned in answer.

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up vote 5 down vote accepted

For some reason, I was not able to edit my original answer. Hence I give a new, elaborated, answer.

Original answer:

Hint: For any $k \in \mathbb{N}$ (arbitrary but fixed), it holds $$ \frac{1}{{1 - x}} - (1 + x + \cdots + x^k ) > \frac{1}{{1 - x}} - (k + 1), $$ for any $0 < x <1$ (real). Now note that $\lim _{x \to 1^ - } \frac{1}{{1 - x}} = \infty $.

Elaborating: Let $S_k (z) = 1 + z + \cdots + z^k$. Since $$ \sum\limits_{k = 0}^\infty {z^k } = \frac{1}{{1 - z}},\;\;|z| < 1, $$ we have that $S_k$ converges pointwise to $1/(1-z)$ for $|z| < 1$. To show that the convergence is not uniform in $|z| < 1$, first note that the set of real numbers $\lbrace 0 < x < 1 \rbrace$ is a subset of $\lbrace |z| < 1 \rbrace$. Thus, given $k \in \mathbb{N}$, arbitrary but fixed, $$ \mathop {\sup }\limits_{|z| < 1} \bigg|\frac{1}{{1 - z}} - S_k (z) \bigg| \ge \mathop {\sup }\limits_{0 < x < 1} \bigg|\frac{1}{{1 - x}} - S_k (x) \bigg|. $$ From $$ \frac{1}{{1 - x}} - S_k (x) > \frac{1}{{1 - x}} - (k + 1), \;\; 0 < x < 1, $$ and $\lim _{x \to 1^ - } \frac{1}{{1 - x}} = \infty $, we get $$ \mathop {\lim }\limits_{x \to 1^ - } \bigg(\frac{1}{{1 - x}} - S_k (x)\bigg) = \infty. $$ Hence also $$ \mathop {\sup }\limits_{0 < x < 1} \bigg|\frac{1}{{1 - x}} - S_k (x) \bigg| = \infty $$ and, in turn, $$ \mathop {\sup }\limits_{|z| < 1} \bigg|\frac{1}{{1 - z}} - S_k (z) \bigg| = \infty. $$ Therefore, since $k \in \mathbb{N}$ is arbitrary, $S_k$ does not converge uniformly to $1/(1-z)$ in $|z| < 1$.

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(I deleted my first answer, which I couldn't edit.) – Shai Covo Jul 12 '11 at 11:19

This is a simple consequence of the fact that each function $S_k:x\mapsto1+x+\cdots+x^k$ is bounded while the limit function $S:x\mapsto1/(1-x)$ is not.

Hence each function $S_k-S$ is unbounded, that is, the sup-norm of $S_k-S$ is infinite, in particular the sequence of the sup-norms does not converge to zero. This last assertion is equivalent to the fact that $(S_k)$ does not converge uniformly to $S$.

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The $n$th partial sum of your series is $S_n(z) = {\displaystyle\sum_{k=0}^n z^k = {1 - z^{n+1} \over 1 - z}}$, using the summation formula for geometric series. The sum of the overall series is ${\displaystyle \sum_{k=0}^{\infty} z^k = {1 \over 1 - z}}$.

To say the series converges uniformly to the limit on the disc is to say that $S_n(z)$ converges uniformly to ${\displaystyle {1 \over 1 - z}}$ on the disc. In other words, you have uniform convergence if for every $\epsilon > 0$ there is an $N$ such that if $n > N$ then for all $z$ with $|z| < 1$ you have $$ |S_n(z) - {1 \over 1 - z}| < \epsilon$$ But $$|S_n(z) - {1 \over 1 - z}| = |{1 - z^{n+1} \over 1 - z} - {1 \over 1 - z}|$$ $$ = |{-z^{n+1} \over 1 - z}|$$ $$= {|z|^{n+1} \over |1 - z|}$$ So the question becomes, is it true that for every $\epsilon > 0$ there is an $N$ such that if $n > N$ then for all $z$ with $|z| < 1$ you have $${|z|^{n+1} \over |1 - z|} < \epsilon $$ But this is clearly false. For any $n$, the limit ${\displaystyle \lim_{x \rightarrow 1^-}{x^{n+1} \over 1 - x} = \infty}$, so there is no $n$ for which ${|z|^{n+1} \over |1 - z|} < \epsilon $ holds for all $z$ with $0 < z < 1$, much less for all $z$ on the unit disc. So uniform convergence does not hold.

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Confine attention to real $x$ in the interval $0<x<1$. Let $$s_n(x)=1+x+x^2+\cdots +x^{n-1}.$$ If we use $s_n(x)$ to approximate the sum, the truncation error is $>x^n$.

Choose a positive $\epsilon$, where for convenience $\epsilon<1$. We want to make the truncation error $<\epsilon$, so we need $$x^n <\epsilon,\qquad \text{or equivalently}\qquad n >\frac{|\ln(\epsilon)|}{|\ln(x)|}.$$ Since $\ln x \to 0$ as $x \to 1^{-}$, the required $n$ grows without bound as $x\to 1^{-}$.

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If I take your wording literally, your difficulty might stem from the fact that you confused the supremum going to zero with the series going to zero pointwise. This is precisely the difference between convergence and uniform convergence. The expression $|g(z)-g_k(z)|$ does go to zero for all $z$ on the open unit disc, but the supremum doesn't. (I'm assuming that by $g(z)$ you mean the series and by $g_k(z)$ its $k$-th partial sum; the question should introduce the notation used if it's not obviously standard.)

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That assumption is correct regarding the notation. I fixed the original question, as I think I wrote it in a confusing fashion. I'm aware of the difference between the two types of convergence, but I'm finding it hard to show that the sup does not equal zero for the expression $|g(z) - g_k(z)|$ – I Love Cake Jul 12 '11 at 9:24
What you changed was just a minor abuse of language -- the main error is still there -- the question is not whether a series goes to zero (it seems the only series that "this series" could refer to is $g(z)-g_k(z)$) but whether the supremum you wrote down in the preceding sentence goes to zero. (Also, in standard math typesetting, "$\sup$" isn't italicized; there's a command \sup that you can use to get the standard roman version.) – joriki Jul 12 '11 at 9:31

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