I want to show that $\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}$ converges on the $disc|z|<1$, diverges at every point $|z|>1$, and converges uniformly on every disc $|z|<=r$, where 0< r < 1.

I get stuck and don't know how to start. Can anyone tell me how to do it? Thank you guys so much!

  • 1
    $\begingroup$ Find a bound for $\sum_{n = m}^{M} z^n/(1 - z^n)$ that depends only on $r$, not on $z$ or $M$, and such that the bound tends to zero when $m$ tends to infinity. For $|z| > 1$, look at the limit of the general term of the series. $\endgroup$ – user49640 Apr 5 '17 at 3:44

This is mostly a result on real numbers more than anything else, and utilizes a general triangle inequality rule. (I mention this as some complex courses do parts in different orders so I am assuming here you have the triangle and reverse triangle inequalities).

For the first part, we wish to show that the series converges for $|z|<1$. Convergence can be delicate, but a good first step is to see if this thing converges absolutely. In real numbers that's done by taking the "positive" value at every iteration but in complex numbers with a bit of thinking you can see that what you are actually doing is finding the modulus of each iteration.

So a sort of heavy hammer to determine if the series must stay a reasonable "size" (i.e. the modulus doesn't can't to infinity) we want to ask about $\left\vert \sum\limits_{n=1}^\infty \dfrac{z^n}{1-z^n} \right\vert$. Unforutnately, we don't technically know this thing make sense (as was pointed out by zhw below) but we can use it as motivation to (formally) consider

$\left\vert \sum\limits_{n=1}^\infty \dfrac{z^n}{1-z^n} \right\vert \leq \sum\limits_{n=1}^\infty \left\vert \dfrac{z^n}{1-z^n} \right\vert \leq \sum\limits_{n=1}^\infty \dfrac{|z|^n}{1-|z|^n} $

Now, if $|z|<1$, then we know for $N$ big enough, $1-|z| > \frac{1}{2}$, so

$\sum\limits_{n=1}^\infty \dfrac{|z|^n}{1-|z|^n} \leq \sum\limits_{n=1}^\infty 2|z|^n$

Where now our result follows from a the standard result on geometric sequences.

Now that we have an idea of why this thing should converge we can run the argument backward for the actual proof (starting with the result on geometric sequences which is known, and work our way back to the original term).

As a side note remember that, just like absolute convergence, if this were to fail (meaning we show that the above goes to infinity) this does not tell us that the original series diverges. All it would tell us is we need more a delicate tool that smacking it with absolute values and triangle inequality (like when you used an alternating series test in calculus on the alternating harmonic to show it converges).

You can follow the same argument (with a little clever use of the reverse triangle inequality) to get your second result. And for the uniform convergence, note that if you have $|z| < r < 1$, then you can take a closed disk of radius slightly larger than r to get a compact set, which gives you an absolute maximum and minimum. You can use these to go for the uniform part.

I'll leave it there so you can try it, but if you're still have trouble post it and I'll try to clarify.

  • $\begingroup$ You don't really want to start with $$\left\vert \sum\limits_{n=1}^\infty \dfrac{z^n}{1-z^n} \right\vert$$ because you are assuming the expression inside the absolute values has meaning, i.e., the series converges. But that is what you are trying to prove. $\endgroup$ – zhw. Apr 5 '17 at 7:12
  • $\begingroup$ You are correct! We don't know that the object is convergent and so writing that term mathematically is begging the question. My original write-up was intended as a motivational 'nudge' to help get parting on the right track conceptually, but I can see how that would have been easily mistaken as an actual proof. I have edited the answer to reflect this. $\endgroup$ – Jason Apr 5 '17 at 12:07

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