Yet another family of hypergeometric sums that has a closed form solution.

Question

Let $m \ge 2$ and $j\ge 0$ be integers. Now, let $0 < z < \frac{(m-1)^{m-1}}{m^m}$ be a real number. Consider a following sum: \begin{equation} {\mathfrak S}^{(m,j)}(z) := \sum\limits_{i=0}^\infty \binom{m \cdot i + j}{i}\cdot z^i \end{equation} By using both my answer to Closed form solutions for a family of hypergeometric sums. and by generalizing the approach from my answer to About the identity $\sum\limits_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z}}$ I have derived the following results: \begin{eqnarray} {\mathfrak S}^{(m,0)}(z) &=& \frac{x \left(1-z x^{m-1}\right)}{1-m z x^{m-1}}\\ {\mathfrak S}^{(m,1)}(z) &=& \frac{m}{m-1} \cdot\frac{ x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}-\frac{1}{m-1}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}\\ {\mathfrak S}^{(m,2)}(z) &=& \frac{m^2}{(m-1)^2}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)} -\frac{(m-2) }{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\ && -\frac{1}{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}} \cdot \frac{((m-1) x+2) }{ x}\\ {\mathfrak S}^{(m,3)}(z) &=& \frac{m^3}{(m-1)^3}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}- \frac{(m-2)(2 m-3)}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\ &&-\frac{m-3}{(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}}\cdot \frac{((m-1) x+2)}{x}+\\ &&-\frac{1}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{3}{m-1}}\cdot \frac{2(m-1)^2 x^2+6(m-1)x+3(m+2)}{x^2} \end{eqnarray} Here $x:=x(z)$ is obtained in the following way. Out of the solutions of the equation: \begin{equation} 1-x+z \cdot x^m=0 \end{equation} we choose the one that is the closest to unity. Now the question is how does the result look like for arbitrary $j \ge 2$.

Answer 1

Hint: The series ${\mathfrak S}^{(m,j)}(z)$ is strongly related with the generalized binomial series $B_m(z)$ \begin{align*} B_m(z)=\sum_{i=0}^\infty\binom{mi+1}{i}\frac{1}{mi+1}z^i \end{align*} defined in (5.58) in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.

The series $B_m(z)$ satisfies the identities (see (5.60), (5.61)): \begin{align*} B_m(z)^j&=\sum_{i=0}^\infty \binom{mi+j}{i}\frac{j}{mi+j}z^i\\ \frac{B_m(z)^j}{1-m+mB_m(z)^{-1}}&=\sum_{i=0}^\infty\binom{mi+j}{i}z^i \end{align*} so that the following holds \begin{align*} \color{blue}{{\mathfrak S}^{(m,j)}(z)=\frac{B_m(z)^j}{1-m+mB_m(z)^{-1}}} \end{align*}

Answer 2

Here I am using the results provided above by Markus Scheuer. Firstly I provide a closed form expression for the quantity $B_m(z)$. We have: \begin{eqnarray} B_m(z)&=& \sum\limits_{i=0}^\infty \binom{m\cdot i}{i} \cdot \underbrace{\frac{1}{(m-1) i+1}}_{\int\limits_0^1 \theta^{(m-1) i} d\theta} \cdot z^i\\ &=& \int\limits_0^1 \frac{x\cdot(1-z\theta^{m-1}\cdot x^{m-1})}{1-m z \theta^{m-1} \cdot x^{m-1}} d\theta \\ &\underbrace{=}_{z \theta^{m-1}=(\xi-1)/\xi^m}& \frac{1}{z^{1/(m-1)} \cdot(m-1)}\int\limits_0^{x-1} \frac{\xi^{1/(m-1)-1}}{(1+\xi)^{m/(m-1)}} d\xi\\ &=& \left( \frac{x-1}{z \cdot x}\right)^{1/(m-1)} = x \end{eqnarray} where $1-x+ z \cdot x^m=0$.

Therefore using the result quoted by Markus Scheuers' we have:

\begin{equation} {\mathfrak S}^{(m,j)}(z) = \frac{(\frac{x-1}{z\cdot x})^{j/(m-1)}}{1-m+m \cdot (\frac{z \cdot x}{x-1})^{1/(m-1)}} = \frac{x^{j+1}}{(1-m) \cdot x+m} \end{equation}

In particular if $m=1,2$ the result reads: \begin{eqnarray} \left\{ {\mathfrak S}^{(m,j)}\right\}_{m=1}^2 = \left\{ (\frac{1}{1-z})^{j+1}, \frac{(\frac{1-\sqrt{1-4 z}}{2 z})^j}{\sqrt{1-4 z}}\right\} \end{eqnarray} as it should be.