Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sorry for taking this from another question, but the second part was never answered, and I'm not sure how to get there. From: Prove the following equation of complex power series.

Show that for $|z| \lt 1$ with $z \in \Bbb C$, we have

$$ \sum_{k=0}^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \frac{z}{1-z} $$

$$ \sum_{k=0}^\infty \frac{2^k{z^2}^k}{1+{z^2}^{k}} = \frac{z}{1-z} $$

My guess is that the second one is obtained by differentiating the first one or something like that, but I can't manage to prove the first one.

share|improve this question
    
I think it would be better to just ask the second question since the first one was already proved in the earlier question. Also are you the same person (math.stackexchange.com/users/60892/user60892)? If so, I can flag the moderators to merge these two accounts. –  user17762 Feb 5 '13 at 8:27
    
not the same person, and I wasn't sure if the first part could be used to simplify the second part –  Moses G Feb 5 '13 at 8:47
    
Yeah, it would have been best to not include the first one, for now we have repeated answers for that (and I even wrote exactly the same answer Marvis wrote before...) –  Mariano Suárez-Alvarez Feb 5 '13 at 8:58

4 Answers 4

There is a simple proof for second identity. Notice: $$1 - z^{2^{N+1}} = (1-z)\prod_{k=0}^N (1 + z^{2^k})$$ Taking logarithm and apply $-z \frac{d}{dz}$ on both side, we get:

$$\frac{2^{N+1} z^{2^{N+1}}}{1 - z^{2^{N+1}}} = \frac{z}{1-z} - \sum_{k=0}^N \frac{2^k z^{2^k}}{1+z^{2^k}}$$ For $|z| < 1$, L.H.S $\to 0$ when $N \to \infty$, we immediately obtain the second identity: $$\sum_{k=0}^{\infty} \frac{2^k z^{2^k}}{1+z^{2^k}} = \frac{z}{1-z}$$

share|improve this answer

For the first identity, you're trying to prove that $$ 0 = \frac{z}{1-z} - \sum_{k=0}^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \lim_{K\to\infty} \bigg( \frac{z}{1-z} - \sum_{k=0}^K\frac{{z^2}^k}{1-{z^2}^{k+1}} \bigg) $$ for $|z|<1$. If you work out the exact value of the expression on the right-hand side for small values of $K$, you get \begin{align*} \frac{z}{1-z} - \sum_{k=0}^0 \frac{{z^2}^k}{1-{z^2}^{k+1}} &= \frac{z}{1-z} - \frac{z}{1-z^2} = \frac{z^2}{1-z^2} \\ \frac{z}{1-z} - \sum_{k=0}^1 \frac{{z^2}^k}{1-{z^2}^{k+1}} &= \frac{z^2}{1-z^2} - \frac{z^2}{1-z^4} = \frac{z^4}{1-z^4} \\ \frac{z}{1-z} - \sum_{k=0}^2 \frac{{z^2}^k}{1-{z^2}^{k+1}} &= \frac{z^4}{1-z^4} - \frac{z^4}{1-z^8} = \frac{z^8}{1-z^8} \end{align*} and so on, suggesting a pattern that successfully be proved by induction. Then you only need to show that $$ \lim_{K\to\infty} \frac{z^{2^{K+1}}}{1-z^{2^{K+1}}} = 0 $$ for $|z|<1$. The second identity can be proved similarly.

share|improve this answer
    
(+1) nice answer. –  Mhenni Benghorbal Feb 5 '13 at 9:12

For the first one: each summand can be developed in a power series, and when you do this, no two terms in the resulting double series are equal, and all powers of $z$ with exponent at least $1$ appear; this is more or less equivalent to the fact that every number has a unique base-$2$ expression. It is easy to see that the double series converges absolutely where it does (just use the Weierstrass $M$-test) and then we can reorganize it as we wish. If you look at what you got, it is precisely the power series development of the right hand side of the equality.

Doing the same —expanding each term into a power series— in the second sum gives us $$\sum_{k,l\geq0}(-1)^l2^kz^{(l+1)2^k}.$$ Again this converges absolutely by another application of the $M$-test, so we can reorganize. If we group all terms which have a $z^a$ for a given $a\in\mathbb Z$, the coefficient we end up with is $$\sum{(-1)^l2^k}$$ with the sum taken over all $k,l\geq0$ such that $(l+1)2^k=a$. It is a fun exercise to show this is always $1$.

share|improve this answer

This answer is inspired by Marvis' answer to the first part of your previous question. Expanding $1/\bigl(1+z^{2^k}\bigr)$ as a geometric series and ignoring convergence issues we get

$$\sum_{k=0}^\infty\frac{2^kz^{2^k}}{1+z^{2^k}}=\sum_{k,\ell\in\mathbb N}2^kz^{2^k}(-1)^\ell z^{\ell2^k}=\sum_{k,\ell\in\mathbb N}2^k(-1)^\ell z^{2^k(\ell+1)}\,.$$

We want to prove that this sum is equal to $\sum_{m=0}^\infty z^m$. For each $m\in\mathbb N$, let $$A_m=\{(k,\ell): 2^k(\ell+1)=m\}\,.$$

Now our objective is to show that $\sum_{(k,\ell)\in A_m}2^k(-1)^\ell=1$ for each $m$. As in Marvis' answer, we use the fact that each $m$ can be written uniquely as $2^r(2s+1)$, with $r,s\in\mathbb N$. If $(k,\ell)\in A_m$, then $2^k(\ell+1)=2^r(2s+1)$, so necessarily $k\leq r$ (because $2s+1$ is odd) and $\ell=2^{r-k}(2s+1)-1$; in particular $\ell$ is odd for $k<r$ and $\ell$ is even for $k=r$. Therefore your sum becomes

$$\sum_{(k,\ell)\in A_m}2^k(-1)^\ell=\sum_{k=0}^r2^k(-1)^{2^{r-k}(2s+1)-1}=2^r(-1)^{\mathrm{even}}+\sum_{k=0}^{r-1}2^k(-1)^{\mathrm{odd}}=1\,.$$

I leave to you to verify that the double series $\sum_{k,\ell\in\mathbb N}2^kz^{2^k(\ell+1)}$ converges absolutely, which in turn validates all the manipulations above.

share|improve this answer
    
This is very beautiful. +1. –  user17762 Feb 5 '13 at 8:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.