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I'm supposed to use Weierstrass $M$-test to prove uniform convergence of this (I think) geometric series $\sum_{n=0}^{\infty} \frac{1}{(z+5)^n}$ on $A=\{|z| \le 3.5\}$. (Indeed $-5 \notin A$.)

I understand this asks me to do:

Let $f_n,f: A \to \mathbb C$, for $n \ge 1$, where $f_n(z)=\sum_{k=0}^n \frac{1}{(z+5)^n}$ and $f(z) = \frac{z+5}{z+4}$. (Indeed $-4 \notin A$.)

Show that

For any $\varepsilon > 0$, there exists $N_{\varepsilon} > 0$ s.t. $|f_n(z)-f(z)| < \varepsilon$ whenever $n > N_{\varepsilon}$ and $z \in A$.

But...what I'm used to understanding is for the set $B=\{|z+5|>1\}$, we have that $\{'f_n: B \to \mathbb C'\}_{n=1}^{\infty}$ (indeed $-5,-4 \notin B$ too) converges pointwise to $'f: B \to \mathbb C'$ on $B$. I know it's a little weird to use the same $f,f_n$ with a different domain unless 1 is a subset of the other, but hear me out:

Questions:

  1. So...what's the relation between $A$ and $B$? Is 1 a subset of the other?

I guess the idea is that $A \subseteq B$, but I'm not sure how to show this. I mean $x^2+y^2 \le \frac{49}{4}$ implies $x^2+5x+25+y^2>1$? Errr...maybe there's some triangle inequality I've missed?

  1. (If neither is a subset of the other, then) is this probably taken to mean that the geometric series is to be shown that it converges uniformly on $A \cap B$?
  • 2.1. Note: I use round brackets because if 1 is a subset of the other, then (i presume it's A and then) $A \cap B = $ the smaller one (w/c i presume is $A$. I mean I don't think it makes sense to enlarge to the set in order to make something non-uniform into uniform.)
Bernard
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BCLC
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1 Answers1

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For $z \in A$ you have using reverse triangle inequality

$$\frac{3}{2} = 5 - \frac{7}{2}\le 5 - \vert -z \vert \le \vert 5 +z \vert,$$

and therefore for any $n \ge 0$

$$\left\vert \frac{1}{(5+z)^n}\right\vert \le \left(\frac{2}{3}\right)^n.$$

As $\sum_{n=0}^\infty$ of the RHS converges, Weierstrass M-test ensures that $\sum_{n=0}^\infty \frac{1}{(z+5)^n}$ converges uniformly on $A$.

  • thanks mathcounterexamples.net! i wish you didn't spoil the problem though – BCLC Oct 04 '21 at 19:02
  • oh neat i was using the reverse triangle inequality awhile ago. I KNEW IT!!! https://math.stackexchange.com/questions/4267659/check-if-frac1z2-frac-1-n-n-1-infty-is-uniform-or-at-least-po/4267765#4267765 – BCLC Oct 04 '21 at 19:03
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    did you mean $|5 - \vert -z \vert| \le |5+z|$ ? i think i did use reverse triangle but couldn't figure out how to get rid of the...(wait...)oooooooohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh ok got it got it. thanks again – BCLC Oct 04 '21 at 19:06
  • oh nice foreshadowing. i see we're proving more than just that $|5+z| >1 1$ – BCLC Oct 04 '21 at 19:12
  • For any $a \in \mathbb R$, $a \le \vert a \vert$… – mathcounterexamples.net Oct 04 '21 at 19:14
  • that's what i meant by 'oooooooohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh ok got it got it. thanks again' ... – BCLC Oct 04 '21 at 19:15
  • mathcounterexamples.net please use spoiler tags to hide the choice for $M$ – BCLC Oct 05 '21 at 00:57
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    @JohnSmithKyon Why do you need this spoiler? The « difficult » part is the first inequality. The spoiler is a simple consequence of it. – mathcounterexamples.net Oct 05 '21 at 05:25
  • The difficulty I find in 2nd part is using the $\frac 3 2$ because initially we're aiming to just prove $A \subseteq B$ so then you might forget that you actually proved the stronger result of $\ge \frac{3}{2}$ instead of just $> 1$ – BCLC Oct 05 '21 at 05:28
  • mathcounterexamples.net please consider helping in the other triangle inequality question: https://math.stackexchange.com/questions/4267659/check-if-frac1z2-frac-1-n-n-1-infty-is-uniform-or-at-least-po – BCLC Oct 06 '21 at 02:00