Watching this video it is "proven" that:
If $\left |{x}\right |<1$ then $$\displaystyle\sum_{n=0}^\infty \dfrac{2^n x^{2^n}}{1+x^{2^n}} = \dfrac{x}{1-x}$$
The "proof" is as follows:
First we observe that as $\left |{x}\right |<1$ then $$\displaystyle \dfrac{1}{1+x^{2^n}} = \sum_{m=0}^\infty (-1)^m x^{m2^n}$$ That way, we can express $$\displaystyle\sum_{n=0}^\infty \dfrac{2^n x^{2^n}}{1+x^{2^n}} = \sum_{n=0}^\infty \sum_{m=0}^\infty (-1)^m 2^n x^{2^n(m+1)}$$ The idea now it to "collapse"this double series into a power series, so we have to regroup the coefficients corresponding to monomials of the same degree.
To do so, we first observe that there is not an independent term, and that for monomials of odd degree, $x^{2k+1}$, we must have $n=0$ and $m=2k$, so the coefficients for odd degree monomials is always $1$.
On the other hand, for even degree monomials, we express the exponent in the unique form $2^kr$ with $k \geq 1$ and $r$ odd, and then observe that
$$2^n(m+1) = 2^k r \Longleftrightarrow{} \left( n=k \wedge m+1=r \right) \vee \left(0 \leq n <k \wedge m+1=2^{k-n} r \right)$$
In that way, the coefficient associated to $x^{2^k r}$ will be
$$\displaystyle \sum_{n=0}^{k-1} (-1)^{2^{k-n} r -1} 2^n + (-1)^{r-1} 2^k = -\sum_{n=0}^{k-1} 2^n + 2^k = 1$$
We have then proven that the coefficient of even degree monomials is also $1$.
Using all this, we finally conclude that
$$\displaystyle \sum_{n=0}^\infty \dfrac{2^n x^{2^n}}{1+x^{2^n}} = \sum_{n=0}^\infty \sum_{m=0}^\infty (-1)^m 2^n x^{2^n(m+1)} = \sum_{k=1}^\infty x^k = \dfrac{x}{1-x}$$
Now, in this "proof" it is implicitly used that we can regroup and rearrange the terms in the double series the way we want, but this is not always possible and it must be justified.
In that way, I have the following conjecture:
Let $\{a_{n,m}\}$ be a double sequence of real numbers and $\{b_{n,m}\}$ a double sequence of non negative integers such that for every $k \geq 0$ the sets
$$\displaystyle T_k=\{(n,m) \in \mathbb{N} \times \mathbb{N}: b_{n,m}=k\}$$ are finite.
Then, if we define $c_k=\sum_{(n,m) \in T_k} a_{n,m}$, it is satisfied that $$\displaystyle \sum_{k=0}^\infty c_k x^k = \sum_{n=0}^\infty \sum_{m=0}^\infty a_{n,m} x^{b_{n,m}}=\sum_{m=0}^\infty \sum_{n=0}^\infty a_{n,m} x^{b_{n,m}}$$ for every $x$ in the disk of convergence of the power series.
I know this conjecture is false in general (for example, taking $a_{n,m}=n-m$ and $b_{n,m}=n+m$), but I've managed to prove that it is true, for example, if we impose that the $a_{n,m}$ are non-negative. The proof can be found here. It is in Spanish, but I can translate it if necessary.
Now, my question is:
- Can we proof the conjecture for a big enough family of double sequences so that it includes the original example? With a new proof or by adjusting the one I give in the link.
- If not, can we formalice the proof given of the example in some way?
- If none of the above is possible, can ww prove the example by some other method?
Thanks in advance for all the answers.