Is there a closed form solution for the sum $\sum\limits_{M=2}^{\infty} \sum\limits_{n=M}^{\infty} \frac{2^{-M}3^{M-n-1}}{(M-n-1)(n+1)^{2}}$?

Question

While working on another problem, this came up as a sub-step: $$ \sum_{M\ =\ 2}^{\infty}\,\,\,\sum_{n\ =\ M}^{\infty}\ {2^{-M}\ 3^{M - n - 1} \over \left(M - n - 1\right)\left(n + 1\right)^{\,2}} $$

Question: Is there a closed form for this sum?

Notes:

1. For anyone that might ask, "why do you think there is a closed form?", I admit I am not sure if there is a closed form or even if I should expect a closed form.
2. For anyone that might ask, "what have you done?", I don't have anything useful to show for this question itself, as I mentioned this is a sub-step/result in a much longer problem, whose immediate relevance to what I am asking is based on the following sum (which in turn is part of an even larger original problem that I can't get into without derailing this question):

Sum[-1/(n + 1)! Binomial[n, k] (k - 1)! (n - k)! Sum[(-1)^(n - k) (1/2^(2 n + 1 - j - k)) Binomial[ 2 n - j - k, n + 1 - k] (1/((j + k - n) 3^(j + k - n))), {j, 2 n + 1 - k - M, 2 n + 1 - k - M} ], {k, n + 2 - M, n}]

3. As to what kind of closed form I am hoping for: It is a combination (i.e. sum) of $\mathrm{constants}\times\pi^{2}$, $\log^{k}(\mathrm{constants})$, $\mathrm{Li}_{2}(\mathrm{constants}), \mathrm{Li}_{3}(\mathrm{constants})$ and $\zeta(3)$ (based on simpler expressions than these I solved for my original problem). By constants, I mean simple known constants like $2,3, 1/2$ (not necessarily those).

4. Simply trying my luck in Mathematica, it is unable to solve and gives expressions for inner sum in terms of HurwitzLerchPhi functions that it cannot then close for the outer sum.

Answer 1

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Let $\mathcal{S}$ denote the sum of the following (convergent) double infinite series:

$$\mathcal{S}:=\sum_{m=2}^{\infty}\sum_{n=m}^{\infty}\frac{2^{-m}3^{m-n-1}}{\left(m-n-1\right)\left(n+1\right)^{2}}.$$

Shifting the initial indices and changing the order of summation, the double sum can be rewritten as

$$\begin{align} \mathcal{S} &=\sum_{m=2}^{\infty}\sum_{n=m}^{\infty}\frac{2^{-m}3^{m-n-1}}{\left(m-n-1\right)\left(n+1\right)^{2}}\\ &=\sum_{k=1}^{\infty}\sum_{n=k+1}^{\infty}\frac{2^{-k-1}3^{k-n}}{\left(k-n\right)\left(n+1\right)^{2}}\\ &=\sum_{k=1}^{\infty}\sum_{j=k}^{\infty}\frac{2^{-k-1}3^{k-j-1}}{\left(k-j-1\right)\left(j+2\right)^{2}}\\ &=\sum_{j=1}^{\infty}\sum_{k=1}^{j}\frac{2^{-k-1}3^{k-j-1}}{\left(k-j-1\right)\left(j+2\right)^{2}}.\\ \end{align}$$

Then, setting $a:=\frac12$ and $b:=\frac23$, we have

$$\begin{align} \mathcal{S} &=\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{2^{-k-1}3^{k-n-1}}{\left(k-n-1\right)\left(n+2\right)^{2}}\\ &=-\sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{2^{n-k+1}}{2^{n+2}3^{n-k+1}\left(n-k+1\right)\left(n+2\right)^{2}}\\ &=-\sum_{n=1}^{\infty}\frac{\left(\frac12\right)^{n+2}}{\left(n+2\right)^{2}}\sum_{k=1}^{n}\frac{\left(\frac23\right)^{n-k+1}}{\left(n-k+1\right)}\\ &=-\sum_{n=1}^{\infty}\frac{a^{n+2}}{\left(n+2\right)^{2}}\sum_{k=1}^{n}\frac{b^{n-k+1}}{\left(n-k+1\right)},\\ \end{align}$$

and we can rewrite the sum as a triple integral as follows:

$$\begin{align} \mathcal{S} &=-\sum_{n=1}^{\infty}\frac{a^{n+2}}{\left(n+2\right)^{2}}\sum_{k=1}^{n}\frac{b^{n-k+1}}{\left(n-k+1\right)}\\ &=-\sum_{n=1}^{\infty}\frac{a^{n+2}}{\left(n+2\right)^{2}}\sum_{k=1}^{n}\int_{0}^{b}\mathrm{d}t\,t^{n-k}\\ &=-\sum_{n=1}^{\infty}\frac{a^{n+2}}{\left(n+2\right)^{2}}\int_{0}^{b}\mathrm{d}t\,\sum_{k=1}^{n}t^{n-k}\\ &=-\sum_{n=1}^{\infty}\frac{a^{n+2}}{\left(n+2\right)^{2}}\int_{0}^{b}\mathrm{d}t\,\frac{1-t^{n}}{1-t}\\ &=-\sum_{n=1}^{\infty}\int_{0}^{b}\mathrm{d}t\,\frac{1-t^{n}}{1-t}\cdot\frac{a^{n+2}}{\left(n+2\right)^{2}}\\ &=-\int_{0}^{b}\mathrm{d}t\,\sum_{n=1}^{\infty}\frac{1-t^{n}}{1-t}\cdot\frac{a^{n+2}}{\left(n+2\right)^{2}}\\ &=-\int_{0}^{b}\mathrm{d}t\,\sum_{n=1}^{\infty}\frac{1-t^{n}}{1-t}\int_{0}^{a}\mathrm{d}x\,\frac{x^{n+1}}{n+2}\\ &=-\int_{0}^{b}\mathrm{d}t\,\sum_{n=1}^{\infty}\frac{1-t^{n}}{1-t}\cdot\frac{1}{x}\int_{0}^{a}\mathrm{d}x\,\frac{x^{n+2}}{n+2}\\ &=-\int_{0}^{b}\mathrm{d}t\,\sum_{n=1}^{\infty}\frac{1-t^{n}}{1-t}\cdot\frac{1}{x}\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,y^{n+1}\\ &=-\int_{0}^{b}\mathrm{d}t\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\sum_{n=1}^{\infty}\frac{y^{n+1}}{x}\cdot\frac{1-t^{n}}{1-t}\\ &=-\int_{0}^{b}\mathrm{d}t\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{y}{x\left(1-t\right)}\sum_{n=1}^{\infty}\left(y^{n}-y^{n}t^{n}\right)\\ &=-\int_{0}^{b}\mathrm{d}t\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{y}{x\left(1-t\right)}\left[\sum_{n=1}^{\infty}y^{n}-\sum_{n=1}^{\infty}y^{n}t^{n}\right]\\ &=-\int_{0}^{b}\mathrm{d}t\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{y}{x\left(1-t\right)}\left[\frac{y}{1-y}-\frac{yt}{1-yt}\right].\\ \end{align}$$

Changing the order of integration, we can reduce the triple integral to a single-variable integral:

$$\begin{align} \mathcal{S} &=-\int_{0}^{b}\mathrm{d}t\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{y}{x\left(1-t\right)}\left[\frac{y}{1-y}-\frac{yt}{1-yt}\right]\\ &=-\int_{0}^{b}\mathrm{d}t\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{y^{2}\left[\left(1-yt\right)-\left(1-y\right)t\right]}{x\left(1-y\right)\left(1-yt\right)\left(1-t\right)}\\ &=-\int_{0}^{b}\mathrm{d}t\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{y^{2}}{x\left(1-y\right)\left(1-yt\right)}\\ &=-\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{0}^{b}\mathrm{d}t\,\frac{y^{2}}{x\left(1-y\right)\left(1-yt\right)}\\ &=-\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{y}{x\left(1-y\right)}\int_{0}^{b}\mathrm{d}t\,\frac{y}{\left(1-yt\right)}\\ &=\int_{0}^{a}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{y\ln{\left(1-by\right)}}{x\left(1-y\right)}\\ &=\int_{0}^{a}\mathrm{d}y\int_{y}^{a}\mathrm{d}x\,\frac{y\ln{\left(1-by\right)}}{x\left(1-y\right)}\\ &=\int_{0}^{a}\mathrm{d}y\,\frac{y\ln{\left(1-by\right)}}{\left(1-y\right)}\int_{y}^{a}\mathrm{d}x\,\frac{1}{x}\\ &=\int_{0}^{a}\mathrm{d}y\,\frac{y\ln{\left(1-by\right)}\left[\ln{\left(a\right)}-\ln{\left(y\right)}\right]}{\left(1-y\right)}\\ &=\int_{0}^{a}\mathrm{d}y\,\frac{y}{y-1}\ln{\left(\frac{y}{a}\right)}\ln{\left(1-by\right)}.\\ \end{align}$$

It is well-known that any integral of the form $\int\mathrm{d}x\,R(x)\ln(a+bx)\ln(c+dx)$, where $R(x)$ is a rational function, can be systematically reduced to trilogarithms, dilogarithms, and elementary functions. I leave finding an actual explicit result to you since it can be quite tedious to do by hand. Have fun. :)

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