# Partial sum formula of following series

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

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

• If you really want "partial sums" you can always do: take the infinite sum as f(x), then make f(xz) then the expansion of f(xz)/(1-z) in terms of z as an OGF gives the individual infinite sums as coefficients. Then you can extract a particular sum by differentiation. Please look at my proof in math.stackexchange.com/questions/400792/… . It's quite common as a "formal" process; i.e. a symbolic process that generates wanted coefficients without things like convergence. Finding your f(x) above doesn't seem impossible, but is harder. – rrogers Feb 20 at 12:42
• @rrogers Can you elaborate a bit more? I would greatly appreciate it. – user3108815 Mar 1 at 18:19
• It's better to refer to "The Method of Coefficients Donatella Merlini, Renzo Sprugnoli, Maria Cecilia Verri" for instance at: pdfs.semanticscholar.org/bdb4/… , It is also on JSTOR. Work your way down to Theorem 2.1 . The whole paper is easy to read and I found it educational. Since you are still interested I will continue to try to find a closed form (or not) for your raw sum; which is required by 1/(1-x) . BTW: above "individual infinite sums" should have been "individual partial sums" (: – rrogers Mar 2 at 21:00
• If you want I will order one of the used books or you can do it. (Bibliomania (especially mathematical) is a sad disease and should have a NIMH psychiatric classification). smile.amazon.com/gp/offer-listing/0486661652 – rrogers Mar 3 at 14:31
• @rrogers is it wrong of me to take advantage of your disease for my own personal gain? Because I am still interested in a solution, but I don't want to pay for that book. – user3108815 Mar 4 at 0:48

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.

• Looks nice but I don't see how the "diagonal" elements are extracted from G(x,y). I also don't see the last conversion. Could we have a reference for it? It almost makes sense but seems a little wrong (to me). – rrogers Mar 20 at 12:40
• Furthermore, some of us don't know what the residue theorem is, if you have a bit of time, I'd appreciate a full answer! – user3108815 Mar 20 at 21:30
• @p4sch it's also not clear where this $w$ is coming from? what is that? – user3108815 Mar 21 at 4:33
• In the last step, I have used the Cauchy integral formula twice times in order to get all diagonal terms. (The Integral gives exactly the $n$-th diagonal term.) See Wikipedia for Cauchy's integral formula. The residue theorem is an extension of Cauchy's theorem. I am not quite sure, if we can real find a nice identity for the last double path-integral. – p4sch Mar 21 at 12:30
• @p4sch how did you compute the sum as you did in the last step? Aren't the denominator's terms real numbers, and the numerator a constant? Wouldn't then that just be the geometric sum formula? Are there things I'm missing here? – user3108815 Mar 21 at 17:09

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...