Partial sum formula of following series

Question

$$\sum_{m \ge 1} \frac{(xy)^m}{(2m)!(1-y^m)}, \quad\text{where }x,y \in \mathbb N$$

I have, to start, J.Jacquelin's answer.

Answer 1

This is a partial answer giving a double path-integral over the function defined in (1).

Let us write for simplicity $$H(x,y) := \sum_{m=1}^\infty \frac{(xy)^m}{(2m)!(1-y^m)}.$$ This sum consits of all diagonal terms of the double series $$G(x,y):=\sum_{m=1}^\infty \frac{x^m}{(2m)!} \sum_{n=1}^\infty \frac{y^n}{1-y^n}.$$ For $x \in \mathbb{C}_{-} := \mathbb{C} \setminus \{ x \in \mathbb{R} : x \le 0 \}$ we have $$\sum_{m=1}^\infty \frac{x^m}{(2m)!} = \cosh(\sqrt{x}) -1$$ where we take the the main branch for the square root. We also have for $y$ with $ 0< \mathrm{Re}(y) < 1$ $$ \sum_{n=1}^\infty \frac{y^n}{1-y^n} = \frac{\psi_y(1) + \log(1-y)}{\log(y)},$$ where $\psi_q$ denotes the q-Polygamma function. Thus $$\tag{1} G(x,y) = \{ \cosh(\sqrt{x}) -1 \} \frac{\psi_y(1) + \log(1-y)}{\log(y)}.$$ Now, we can apply the Cauchy integral formula to get \begin{align} H(x,y) &= \frac{1}{2\pi i} \sum_{n=1}^\infty \int_{ \partial B_\varepsilon(x)} \int_{\partial B_\varepsilon(y)} \frac{G(z,w)}{(z-x)^n (w-y)^n} \, dw dz \\ &= \frac{1}{2\pi i} \int_{\partial B_\varepsilon(x)} \int_{\partial B_\varepsilon(y)} \frac{G(z,w)}{(z- x)(w-y)-1} \, dw dz. \end{align} At this point, one may use the residue theorem.

Answer 2

To try to answer my own question, here's something I started. Be warned, it ends in a tautology, but it may help spark some thought. All variables below are assumed $\in \mathbb N$ unless stated otherwise.

We start with something similar: $$\sum_{m=1}^\infty \frac{x^m}{m!} = e^{x}$$ $$\sum_{m=1}^\infty \frac{x^m(1-y^m)}{m!(1-y^m)} = e^{x}$$ $$\sum_{m=1}^\infty \frac{x^m}{m!(1-y^m)} - \sum_{m=1}^\infty \frac{x^my^m}{m!(1-y^m)}= e^{x}$$ Now we focus on the first term of the LHS. From the CDF of the Poisson distribution (see chart), we have (if we assume some variable $k \in \mathbb N$ and ommit results for $k=0$) : $$e^{-x}\sum_{m=1}^k \frac{x^m}{m!} = \frac{\Gamma(k+1,\ x)}{k!}$$ $$e^{-x}\sum_{m=1}^{k-1} \frac{x^m}{m!} = \frac{\Gamma(k,\ x)}{(k-1)!}$$

The numerators of the RHS's of the above two lines are the upper incomplete gamma function. See relevant information about the upper incomplete gamma function, we'll use it again later

It follows then that: $$\frac{1}{1-y^k} \sum_{m=1}^k \frac{x^m}{m!} = \frac{1}{1-y^k} \frac{e^x\ \Gamma(k+1,\ x)}{k!}$$ $$\frac{1}{1-y^k} \sum_{m=1}^k \frac{x^m}{m!} - \frac{1}{1-y^k} \sum_{m=1}^{k-1} \frac{x^m}{m!} = \frac{1}{1-y^k} \frac{e^x\ \Gamma(k+1,\ x)}{k!} - \frac{1}{1-y^k} \frac{e^x\ \Gamma(k,\ x)}{(k-1)!} $$ $$\sum_{k=1}^\infty \frac{x^k}{k!(1-y^k)} = \sum_{k=1}^\infty \frac{e^x\ \Gamma(k+1,\ x)}{k!(1-y^k)} - \sum_{k=1}^\infty \frac{e^x\ \Gamma(k,\ x)}{(k-1)!(1-y^k)}$$

The LHS of the above equality should look familiar. Start with the first term of the RHS of the last equality:

$$\sum_{k=1}^\infty \frac{e^x\ \Gamma(k+1,\ x)}{k!(1-y^k)} = \sum_{k=1}^\infty \frac{e^x\ (\Gamma(k+1) - \gamma(k+1,\ x))}{k!(1-y^k)}$$ $$= \sum_{k=1}^\infty \frac{e^x\ \Gamma(k+1)}{k!(1-y^k)} - \sum_{k=1}^\infty e^x \frac{\gamma(k+1,\ x)}{k!(1-y^k)} $$

The lower case gamma is the "lower incomplete gamma function", from the article above on the upper incomplete gamma function. The substitution in the last equality comes from a holomorphic extension cited in the article. The ".gov" site (NIST Digital Library of Mathematical Functions) hosting the quoted result may not be accessible, which is why I "cite" the Wikipedia articles.

$$\sum_{k=1}^\infty \frac{e^x\ \Gamma(k+1)}{k!(1-y^k)} = \sum_{k=1}^\infty \frac{e^x}{1-y^k} = e^x \sum_{k=1}^\infty \frac{1}{1-y^k} = e^x (\frac{\psi_\frac{1}{y}(1)-\ln(y-1)+\ln\frac{1}{y}}{\ln y})$$

$\psi_q(a)$ where $(a,q) \in \mathbb R$ is known as the the q-digamma function (see Wolfram Alpha, Wikipedia, the other answers, etc). A closed form of the RHS is provided above, you can use Wolfram Alpha (Mathematica really) to verify. Onward. There are a few ways to tackle the next part. I don't know which of them will work, so maybe we try the simplest ones first.

$$\sum_{k=1}^\infty e^x\frac{\gamma(k+1,\ x)}{k!(1-y^k)} = \sum_{k=1}^\infty e^x\frac{\sum_{i=0}^\infty \frac{(-1)^i\ (x)^{k+1+i}}{i!(k+1+i)}}{k!(1-y^k)} =\ ...$$

The series representation substitution comes from the Wikipedia article on the Incomplete Gamma Function, specifically the Lower Incomplete Gamma Function section. Here we will cheat; manually input values of $k$ (e.g. increment value by one) and examine the expanded forms of the outputs. These forms are then summed for each $k$. Do this until you can convince yourself that you can and should prove through induction the pattern you observe. We may need Mathematica or some other computational tool in addition to some induction to deduce:

$$...\ = \sum_{k=1}^\infty \frac{e^x}{1-y^k} - \sum_{k=1}^\infty \frac{1}{1-y^k}-\sum_{k=1}^\infty (\frac{x^{k}}{k!} \sum_{i=k}^\infty \frac{1}{1-y^i})$$

The last term's sum looks particularly nasty, and in truth it is. We can make a few observations and simplifications, including that the last term of the RHS is just:

$$-\sum_{k=1}^\infty (\frac{x^{k}}{k!} \sum_{i=k}^\infty \frac{1}{1-y^i}) = -\sum_{k=1}^\infty (\frac{x^{k}}{k!} (\sum_{i=1}^\infty \frac{1}{1-y^i} - \sum_{i=1}^{k-1} \frac{1}{1-y^i}))$$ $$= -\sum_{k=1}^\infty \frac{x^{k}}{k!}(\sum_{i=1}^\infty \frac{1}{1-y^i}+ \frac{\psi_y(k)}{\ln y}-(k-1)-\frac{\psi_y(1)}{\ln y})$$ After doing the distribution of the $\frac{x^k}{k!}$ across the terms and noting that $$\sum_{k=1}^\infty (\frac{x^{k}}{k!} \sum_{i=1}^\infty \frac{1}{1-y^i}) =e^x \sum_{i=1}^\infty \frac{1}{1-y^i} = e^x \sum_{k=1}^\infty \frac{1}{1-y^k} = e^x (\frac{\psi_\frac{1}{y}(1)-\ln(y-1)+\ln\frac{1}{y}}{\ln y})$$ you'll find the new sums equal to the following:$$-\sum_{k=1}^\infty (\frac{x^{k}}{k!} \sum_{i=k}^\infty \frac{1}{1-y^i}) = - \sum_{k=1}^\infty \frac{e^x}{1-y^k}- \frac{1}{\ln y} \biggr (\sum_{k=1}^\infty \frac{x^{k}}{k!}\psi_y(k)\biggl ) +e^x(x-1)+1+\frac{e^x\ \psi_y(1)}{\ln y}$$

I can immediately see a large problem with the sum using $\psi_y(k)$. Press on though, by taking a step back to perform the substitutions to simplify the work a bit:

$$\sum_{k=1}^\infty e^x\frac{\gamma(k+1,\ x)}{k!(1-y^k)} = \sum_{k=1}^\infty \frac{e^x}{1-y^k} - \sum_{k=1}^\infty \frac{1}{1-y^k}-\sum_{k=1}^\infty (\frac{x^{k}}{k!} \sum_{i=k}^\infty \frac{1}{1-y^i})$$ $$ = - \sum_{k=1}^\infty \frac{1}{1-y^k}- \frac{1}{\ln y} \biggr (\sum_{k=1}^\infty \frac{x^{k}}{k!}\psi_y(k)\biggl ) +e^x(x-1)+1+\frac{e^x\ \psi_y(1)}{\ln y}$$

Continue by pausing this particular expansion and "simplify" a previous equality's terms. From: $$\sum_{k=1}^\infty \frac{x^k}{k!(1-y^k)} = \sum_{k=1}^\infty \frac{e^x\ \Gamma(k+1,\ x)}{k!(1-y^k)} - \sum_{k=1}^\infty \frac{e^x\ \Gamma(k,\ x)}{(k-1)!(1-y^k)}$$

We now focus on the expanded second term of the RHS, and use the same tricks as before: $$- \sum_{k=1}^\infty \frac{e^x\ \Gamma(k,\ x)}{(k-1)!(1-y^k)} = -\sum_{k=1}^\infty \frac{e^x\ \Gamma(k)}{(k-1)!(1-y^k)} + \sum_{k=1}^\infty e^x \frac{\gamma(k,\ x)}{(k-1)!(1-y^k)}$$ $$= -\sum_{k=1}^\infty \frac{e^x}{1-y^k} + \sum_{k=1}^\infty e^x\frac{\sum_{i=0}^\infty \frac{(-1)^i\ (x)^{k+i}}{i!(k+i)}}{(k-1)!(1-y^k)}$$ $$\ =-\sum_{k=1}^\infty \frac{e^x}{1-y^k} +\sum_{k=1}^\infty \frac{e^x}{1-y^k} - \sum_{k=1}^\infty \frac{1}{1-y^k}-\sum_{k=2}^\infty \biggr (\frac{x^{k-1}}{(k-1)!} \sum_{i=k}^\infty \frac{1}{1-y^i}\biggl )$$ $$\ =- \sum_{k=1}^\infty \frac{1}{1-y^k}- \sum_{k=1}^\infty \frac{e^x-1}{1-y^k}- \frac{1}{\ln y} \biggr (\sum_{k=2}^\infty \frac{x^{k-1}}{(k-1)!}\psi_y(k)\biggl ) +xe^x+\frac{(e^x-1)\ \psi_y(1)}{\ln y}$$ $$\ =- \sum_{k=1}^\infty \frac{e^x}{1-y^k}- \frac{1}{\ln y} \biggr (\sum_{k=2}^\infty \frac{x^{k-1}}{(k-1)!}\psi_y(k)\biggl ) +xe^x+\frac{(e^x-1)\ \psi_y(1)}{\ln y}$$

Now we can start combining our results! First, we have to note a few things: we pulled a trick in the 2nd to last line, on the last term of the RHS. Instead of writing: $$\sum_{k=1}^\infty \biggr (\frac{x^{k}}{k!} \sum_{i=k+1}^\infty \frac{1}{1-y^i}\biggl )$$ We wrote it as: $$\sum_{k=2}^\infty \biggr (\frac{x^{k-1}}{(k-1)!} \sum_{i=k}^\infty \frac{1}{1-y^i}\biggl )$$

I entreat you to prove to me they are not equivalent, given our stated variable constraints. Seriously, please do. Next, we never actually calculated $$\sum_{k=1}^\infty \frac{e^x\ \Gamma(k+1,\ x)}{k!(1-y^k)}$$ by combining our previous results of the expansions of the two terms in its own expansion (the one we got via holomorphic extension from the article, remember?).

$$\sum_{k=1}^\infty \frac{e^x\ \Gamma(k+1,\ x)}{k!(1-y^k)} = \sum_{k=1}^\infty \frac{e^x\ \Gamma(k+1)}{k!(1-y^k)} - \sum_{k=1}^\infty e^x \frac{\gamma(k+1,\ x)}{k!(1-y^k)}$$ $$ = \sum_{k=1}^\infty \frac{e^x}{1-y^k}+ \sum_{k=1}^\infty \frac{1}{1-y^k}+ \frac{1}{\ln y} \biggr (\sum_{k=1}^\infty \frac{x^{k}}{k!}\psi_y(k)\biggl ) -e^x(x-1)-1-\frac{e^x\ \psi_y(1)}{\ln y}$$ Combine the above result with one of our prior results: $$- \sum_{k=1}^\infty \frac{e^x\ \Gamma(k,\ x)}{(k-1)!(1-y^k)}=- \sum_{k=1}^\infty \frac{e^x}{1-y^k}- \frac{1}{\ln y} \biggr (\sum_{k=2}^\infty \frac{x^{k-1}}{(k-1)!}\psi_y(k)\biggl ) +xe^x+\frac{(e^x-1)\ \psi_y(1)}{\ln y}$$ To obtain: $$\sum_{k=1}^\infty \frac{x^k}{k!(1-y^k)} = \sum_{k=1}^\infty \frac{e^x\ \Gamma(k+1,\ x)}{k!(1-y^k)} - \sum_{k=1}^\infty \frac{e^x\ \Gamma(k,\ x)}{(k-1)!(1-y^k)}$$ $$= \sum_{k=1}^\infty \frac{1}{1-y^k}+ \frac{1}{\ln y} \biggr (\sum_{k=1}^\infty \frac{x^{k}}{k!}\psi_y(k)\biggl ) -e^x(x-1)-1-\frac{e^x\ \psi_y(1)}{\ln y}- \frac{1}{\ln y} \biggr (\sum_{k=2}^\infty \frac{x^{k-1}}{(k-1)!}\psi_y(k)\biggl ) +xe^x+\frac{(e^x-1)\ \psi_y(1)}{\ln y}$$ $$=\sum_{k=1}^\infty \frac{1}{1-y^k}+ \frac{1}{\ln y} \biggr (\sum_{k=1}^\infty \frac{x^{k}}{k!}\psi_y(k)\biggl ) - \frac{1}{\ln y} \biggr (\sum_{k=2}^\infty \frac{x^{k-1}}{(k-1)!}\psi_y(k)\biggl ) +e^x-1 -\frac{\psi_y(1)}{\ln y}$$

The last "trick" is to take care of those pesky q-digamma function sums. You may use Wolfram Alpha to verify. Observe: $$- \frac{1}{\ln y} \biggr (\sum_{k=2}^\infty \frac{x^{k-1}}{(k-1)!}\psi_y(k)\biggl ) = - \frac{1}{\ln y} \biggr (\sum_{k=1}^\infty \frac{x^{k}}{k!}\psi_y(k+1)\biggl ) $$ $$\frac{1}{\ln y} \biggr (\sum_{k=1}^\infty \frac{x^{k}}{k!}\psi_y(k)\biggl ) - \frac{1}{\ln y} \biggr (\sum_{k=2}^\infty \frac{x^{k-1}}{(k-1)!}\psi_y(k)\biggl ) = \frac{1}{\ln y} \biggr (\sum_{k=1}^\infty \frac{x^{k}}{k!}(\psi_y(k)-\psi_y(k+1))\biggl )$$ $$= -\sum_{k=1}^\infty \frac{x^ky^k}{k!(y^k-1)}$$ Reinsert this new "simplification" into the last main result:

$$\sum_{k=1}^\infty \frac{x^k}{k!(1-y^k)} =\sum_{k=1}^\infty \frac{1}{1-y^k}- \sum_{k=1}^\infty \frac{x^ky^k}{k!(y^k-1)} +e^x-1 -\frac{\psi_y(1)}{\ln y}$$

Finally, we return to one of our earliest results (employ a "change of variables" and use $k$ instead of $m$ like we did above): $$\sum_{k=1}^\infty \frac{x^k}{k!(1-y^k)} - \sum_{k=1}^\infty \frac{x^ky^k}{k!(1-y^k)}= e^{x}$$ Make the appropriate substitution: $$\sum_{k=1}^\infty \frac{1}{1-y^k}- \sum_{k=1}^\infty \frac{x^ky^k}{k!(y^k-1)} +e^x-1 -\frac{\psi_y(1)}{\ln y} - \sum_{k=1}^\infty \frac{x^ky^k}{k!(1-y^k)}= e^{x}$$

Observe another "simplification": $$ \sum_{k=1}^\infty \frac{x^ky^k}{k!(y^k-1)} = - \sum_{k=1}^\infty \frac{x^ky^k}{k!(1-y^k)}$$

Make the substitution: $$\sum_{k=1}^\infty \frac{1}{1-y^k}+ \sum_{k=1}^\infty \frac{x^ky^k}{k!(1-y^k)} +e^x-1 -\frac{\psi_y(1)}{\ln y} - \sum_{k=1}^\infty \frac{x^ky^k}{k!(1-y^k)}= e^{x}$$

$$\sum_{k=1}^\infty \frac{1}{1-y^k} -1 -\frac{\psi_y(1)}{\ln y} = 0$$ Which is not helpful at all! Somewhere we got caught in a tautology, will contemplate how to fix...