Show $1+\frac{8q}{1-q}+\frac{16q^2}{1+q^2}+\frac{24q^3}{1-q^3}+\dots=1+\frac{8q}{(1-q)^2}+\frac{8q^2}{(1+q^2)^2}+\frac{8q^3}{(1-q^3)^2}+\dots$.

Question

Show that $$1+\frac{8q}{1-q}+\frac{16q^2}{1+q^2}+\frac{24q^3}{1-q^3}+\dots=1+\frac{8q}{(1-q)^2}+\frac{8q^2}{(1+q^2)^2}+\frac{8q^3}{(1-q^3)^2}+\dots$$ where $|q|<1$ (q can be complex number).

The hint is to convert the left side to a double series.

One can see that the nominators on the left side is double, triple, ... of those on the right side, while denominators on the right side are square of those on the left side. Both sides have 1, which seems redundant.

I'm not sure how to proceed from these observations. Any advice will be helpful. I can try to solve it with some suggestion and if I still can't solve it I will then explain what puzzles me. If anyone gives the answer I guess I will look at only part of it and work from that.

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Edits:

I post my answer inspired by the comments and some other thoughts about the question below.

Is there any other ways to solve it?

Answer 1

I guess one way is to try to decompose the series to product of two absolutely convergent series, another way is to prove in general what makes $\sum\frac{na_n}{b_n}=\sum\frac{a_n}{(b_n)^2}$; or I can check if the equation is valid for all q and see if I can get some ideas.

I try to decompose the series as $\sum \frac{n(\sqrt{q})^n(\sqrt{q})^n}{1+(i\sqrt{q})^{2n}}=\sum \frac{n(\sqrt{q})^n(\sqrt{q})^n}{1-(i\sqrt{q})^{2n}i^2}=\sum\frac{n(\sqrt{q})^n(\sqrt{q})^n}{(1-(i\sqrt{q})^{n}i)(1+(i\sqrt{q})^{n}i)},$ which is a product of two absolutely convergent series. Is it possible to proceed from that?

I guess a problem I encounter here is that I can't introduce another index, or even make the series product of two (perhaps absolutely convergent) series with independent indexes.

(Well, perhaps I should make it more explicit that it's not exactly $$1+\frac{8q}{1+q}+\frac{16q^2}{1-q^2}+\frac{24q^3}{1+q^3}+\dots,$$

which is very similar to the above series. One may not express the denominators elegantly, which I guess is something one naturally pursues, but at least in this case it doesn’t hinder us from finding a nice solution.)

However, the two actually work the same way. (Note that $1/(1-a)$, instead of $1/(1-a^n)$, expands to $1+a+a^2+...$ this is exactly where I get trapped.)

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Here is the solution I get by far: $$\frac{q}{1+q}+\frac{2q^2}{1-q^2}+\frac{3q^3}{1+q^3}+\dots =\sum \frac{nq^n}{1-(-)^{n+1}q^n}\\ =\sum_{n=1} nq^n\sum_{m=0} ((-)^{n+1}q^n)^m\\ =\sum_{m=0}\sum_{n=1} (-)^{m}n[(-)^{m}q^{m+1}]^n,$$

which, for $\sum_{n=1} nt^n=\frac{t}{(1-t)^2}$ , equals $$\sum_{m=0} (-)^{m}\frac{(-)^{m}q^{m+1}}{(1-(-)^{m}q^{m+1})^2}=\sum_{m=0} \frac{q^{m+1}}{(1-(-)^{m}q^{m+1})^2},$$

and so we get the result.

In a word, $\frac{1}{1-t}=\sum_{m=0} t^m, \sum_{n=1} nt^n=\frac{t}{(1-t)^2}$, this is how we go from the left side to the right.

The double series is like:

$1\ \ \ \ \ \ (1\ +0\ \ \ \ \ \ \ \ +0^2+...\ \ \ \ \ )$

$8q\ \ \ \ (1\ +q\ \ \ \ \ \ \ \ +q^2+...\ \ \ \ \ )$

$16q^2(1\ +(-q^2)\ +(-q^2)^2...)$

And for the series converges absolutely, the row sum (left side) and the column sum (right side) is the same, which completes the proof.

This example illustrates a condition for switching the order of summing indexes, as well as how to recognize patterns. Also, $nq^n$ and $q^n$ are exactly the two absolute convergent series that I was looking for, though I got the latter not by breaking each item on the left to a prduct, but by using Taylor expansion. From that one can see matrix, indexing, (double) series, (Taylor) expansion are much related.

PS: a hindsight, one may view the problem from another way. The sum of nominator $8nq^n$ in the series is obviously convergent, while the denominator is not, so we may regard it as a limit of infinite series, i.e. a sum, instead of as an item in the series.

Answer 2

These series are not equal for all $|q|<1$. Here's an explanation.

First, let $$A(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n},\qquad |bq|<1$$ and $$B(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{(1-bq^n)^2},\qquad |bq|<1.$$ Let $$F_1(q)=A(-2,-1;-q)=\frac{2q}{1-q}+\frac{4q}{1+q^2}+\frac{8q^3}{1-q^3}+...$$ and $$F_2(q)=B(-1,-1;-q)=\frac{q}{(1-q)^2}+\frac{q^2}{(1+q^2)^2}+\frac{q^3}{(1-q^3)^2}+...$$

You assert that $$1+4F_1(q)=1+8F_2(q).\tag{*}$$ This is incorrect. Below is an explanation.

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First of all, define the functions $\phi_n,\sigma_n:\Bbb R\times\Bbb R\to \Bbb R$ to satisfy $$\begin{align} A(a,b;q)&=\sum_{n\ge1}\phi_n(a,b)q^n\\ B(a,b;q)&=\sum_{n\ge1}\sigma_n(a,b)q^n. \end{align}$$ Then we can see that $$\begin{align} A(a,b;q)&=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n}\\ &=\frac1b\sum_{n\ge1}a^n\frac{bq^n}{1-bq^n}\\ &=\frac1b\sum_{n\ge1}a^n\sum_{k\ge1}b^kq^{nk}\\ &=\frac1b\sum_{n\ge1}\sum_{k\ge1}a^nb^kq^{nk}\\ &=\frac1b\sum_{m\ge1}q^m\sum_{nk=m}a^nb^k\\ &=\sum_{m\ge1}q^m\sum_{d|n}a^db^{m/d-1}. \end{align}$$ Thus $$\phi_n(a,b)=\sum_{d|n}a^db^{n/d-1}.$$ Then, preform the index shift $d=n/j$ for $j|n$: $$\phi_n(a,b)=\sum_{j|n}a^{n/j}b^{j-1}=\frac{a}{b}\sum_{j|n}b^ja^{n/j-1}=\frac{a}{b}\phi_n(b,a).\tag1$$

Then take $\partial_b=\frac{\partial}{\partial b}$ of $A(a/q,b;q)$: $$\begin{align} \partial_b A(a/q,b;q)&=\sum_{n\ge1}\partial_b\frac{a^n}{1-bq^n}\\ &=\sum_{n\ge1}\frac{(aq)^n}{(1-bq^n)^2}\\ &=B(a,b;q). \end{align}$$ Then from $(1)$, we have that $A(a,b;q)=\frac{a}{b}A(b,a;q)$, so that $$\begin{align} A(a/q,b;q)&=\frac{a}{bq}A(b,a/q;q)\\ &=a\sum_{n\ge1}\frac{(bq)^{n-1}}{1-aq^{n-1}}\\ &=\frac{a}{1-a}+a\sum_{n\ge1}\frac{(bq)^{n}}{1-aq^{n}}\\ &=\frac{a}{1-a}+bA(a,b;q). \end{align}$$ Thus, $$\begin{align} B(a,b;q)&=\partial_b A(a/q,b;q)\\ &=\partial_b \left\{\frac{a}{1-a}+bA(a,b;q)\right\}\\ &=\partial_b bA(a,b;q)\\ &=A(a,b;q)+b\partial_b A(a,b;q)\\ &=\sum_{n\ge1}q^n\phi_n(a,b)+\sum_{n\ge1}q^nb\partial_b \phi_n(a,b)\\ &=\sum_{n\ge1}q^n\left\{\sum_{d|n}a^db^{n/d-1}+b\sum_{d|n}\left(\tfrac{n}{d}-1\right)a^db^{n/d-2}\right\}\\ &=\sum_{n\ge1}q^n\sum_{d|n}\tfrac{n}{d}a^db^{n/d-1}. \end{align}$$ Hence $$\sigma_n(a,b)=\sum_{d|n}\tfrac{n}{d}a^db^{n/d-1}.$$

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Since $F_1(q)=A(-2,-1;-q)$ and $F_2(q)=B(-1,-1;-q)$, we have $$F_1(q)=\sum_{n\ge1}q^n\cdot\left((-1)^n\phi_n(-2,-1)\right)=\sum_{n\ge1}p_1(n)q^n,$$ and $$F_2(q)=\sum_{n\ge1}q^n\cdot\left((-1)^n\sigma_n(-1,-1)\right)=\sum_{n\ge1}p_2(n)q^n.$$ Then your claim $(*)$ is equivalent to $p_1(n)=2p_2(n)$, or $$\sum_{d|n}(-1)^{d+n/d-1}2^d=2\sum_{d|n}(-1)^{d+n/d-1}\frac{n}{d}.\tag{*'}$$ So $(*)$ is true if and only if $(*')$ is true.

To check this, let $$f(n)=\sum_{d|n}(-1)^{d+n/d-1}(2^d-2n/d).$$ $(*)$ Is true if and only if $f(n)$ is identical to $0$.

It is quite easy to show that $f(3)=-(2-6)-(2^3-2)=-2$, so $f(n)$ is not identical to $0$ and your claim is false.