Complex Analysis by Elias M. Stein Chap 1, Exercise 21

Question

Show that if $|z|<1$ and m is a positive integer, one has $$\frac{z}{1-z^{2}}+\frac{z^{2}}{1-z^{4}}+...+\frac{z^{2^{n}}}{1-z^{2^{n+1}}}+…=\frac{z}{1-z}$$ and $$\frac{z}{1+z}+\frac{2z^{2}}{1+z^{2}}+...+\frac{2^{k}z^{2^{k}}}{1+z^{2^{k}}}+…=\frac{z}{1-z}.$$ Justify any change in the order of summation. [Hint: Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^{2}+...+2^{k}.$]

I tried to prove: $$z^{m}=\sum^{\infty}_{n=2^{m-1}}z^{2^{m}n}$$

which I obtained after several steps of converting fractions to series, but I don't know how to use the hint.

Not the solution they want, but if you call your sum $S$ then $ z^{-1}S = \frac{d}{dz} \sum_{n \ge 0}\ln\left(1+z^{2^n}\right). $ — NoName, Sep 08 '19 at 06:49
@JohnHe Since this is a duplicate I moved my answer to https://math.stackexchange.com/a/3348046/299698 — Robert Z, Sep 08 '19 at 07:19

score 2 · Answer 1 · edited Sep 09 '19 at 11:21

2

Concerning the first sum, note that:

$\displaystyle \frac z{1-z^2}=z+z^3+z^5+z^7+ \cdots$ (the exponents are the odd natural numbers);
$\displaystyle \frac{z^2}{1-z^4}=z^2+z^6+z^{10}+z^{14}+ \cdots$ (the exponents are the even natural numbers which are not a multiple of $4$);
$\displaystyle \frac{z^4}{1-z^8}=z^4+z^{12}+z^{20}+z^{28}+ \cdots$ (the exponents are the natural numbers which are a multiple of $4$ but not a multiple of $8$)

and so on. So, $\displaystyle\sum_{k=0}^\infty\frac{z^{2^k}}{1-z^{2^{k+1}}}$ is the sum of all $z^k$'s, which means that it is equal to $\displaystyle\frac z{1-z}$. Of course, the “So” from the previous sentence must be justified.

Now, try the same approach for the other sum.

edited Sep 09 '19 at 11:21

Hussain-Alqatari

5,065
2
13
39

answered Sep 08 '19 at 06:41

José Carlos Santos

427,504

Great. But it's not hard to show $0=1$ by a formally identical argument; you might mention why it's ok here to take $\sum_j a_j=\sum_n\sum_{j\in S_n}a_j$. – David C. Ullrich Sep 08 '19 at 13:38
The OP claimed not to “know how to use the hint”. My answer explains what the problem has to do with changing the order of summation. But, yes, the OP should now explain why that can be done here. – José Carlos Santos Sep 08 '19 at 13:42
ok............. – David C. Ullrich Sep 08 '19 at 13:44

Complex Analysis by Elias M. Stein Chap 1, Exercise 21

1 Answers1