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

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.

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 $$\sigma_n(a,b)=\sum_{d|n}\tfrac{n}{d}a^db^{n/d-1}.$$

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.

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.)

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.