Infinite sum of falling factorial quotients.

Question

Computing the infinite sum of falling factorial quotients in Mathematica I've got the result: $$ \sum_{k=0}^\infty\frac{z_1^\underline{k}}{(-z_2)^\underline{k}}=\frac{\Gamma(z_2)\Gamma(z_1+z_2-1)}{\Gamma(z_2-1)\Gamma(z_1+z_2)}. $$ How to prove the result? What are the necessary and sufficient conditions for convergence of the series, assuming that both $z_1$ and $z_2$ are not integer? Any hint is appreciated.

Answer 1

Evaluation

$$\newcommand{\Re}{\operatorname{Re}} \begin{align} \sum_{k=0}^\infty\frac{z_1^\underline{k}}{(-z_2)^\underline{k}} &=\sum_{k=0}^\infty\frac{(k-z_1-1)^{\underline{k}}}{(k+z_2-1)^{\underline{k}}}\tag1\\ &=\sum_{k=0}^\infty\frac{\Gamma(k-z_1)}{\Gamma(-z_1)}\frac{\Gamma(z_2)}{\Gamma(k+z_2)}\tag2\\ &=\frac{\Gamma(z_2)}{\Gamma(-z_1)\Gamma(z_1+z_2)}\sum_{k=0}^\infty\frac{\Gamma(k-z_1)\Gamma(z_1+z_2)}{\Gamma(k+z_2)}\tag3\\ &=\frac{\Gamma(z_2)}{\Gamma(-z_1)\Gamma(z_1+z_2)}\sum_{k=0}^\infty\int_0^\infty\frac{t^{z_1+z_2-1}}{(1+t)^{k+z_2}}\,\mathrm{d}t\tag4\\ &=\frac{\Gamma(z_2)}{\Gamma(-z_1)\Gamma(z_1+z_2)}\int_0^\infty\frac{t^{z_1+z_2-2}}{(1+t)^{z_2-1}}\,\mathrm{d}t\tag5\\[6pt] &=\frac{\Gamma(z_2)}{\Gamma(-z_1)\Gamma(z_1+z_2)}\frac{\Gamma(z_1+z_2-1)\Gamma(-z_1)}{\Gamma(z_2-1)}\tag6\\[3pt] &=\bbox[5px,border:2px solid #C0A000]{\frac{z_2-1}{z_1+z_2-1}}\tag7 \end{align} $$ Explanation:
$(1)$: $z^{\underline{k}}=(-1)^k(k-z-1)^{\underline{k}}$
$(2)$: write the falling factorial using the Gamma Function
$(3)$: algebraic manipulation
$(4)$: apply the Beta Function integral
$(5)$: sum the geometric series
$(6)$: apply the Beta Function integral, which converges for $\Re(z_1)\lt0$ and $\Re(z_1+z_2)\gt1$
$(7)$: $\Gamma(z+1)=z\,\Gamma(z)$

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Convergence

Note that as $k\to\infty$, $$ \begin{align} \frac{z_1^{\underline{k}}}{(-z_2)^{\underline{k}}} &=\frac{(k-z_1-1)^{\underline{k}}}{(k+z_2-1)^{\underline{k}}}\\ &=\frac{\Gamma(k-z_1)}{\Gamma(-z_1)}\frac{\Gamma(z_2)}{\Gamma(k+z_2)}\\ &\sim\frac{\Gamma(z_2)}{\Gamma(-z_1)}k^{-z_1-z_2}\tag8 \end{align} $$ Therefore, as long as $z_2$ is not a non-positive integer (so that $(-z_2)^{\underline{k}}$ never vanishes), the series will converge for $\Re(z_1+z_2)\gt1$. By analytic continuation, $(7)$ will hold for $\Re(z_1+z_2)\gt1$.

Answer 2

Let $$S = \sum_{k = 0}^\infty \frac{(z_1)_k}{(-z_2)_k},$$ where $(x)_k = x^{\underline{k}}$ is the falling factorial.

We will make use of the following property for the falling factorial $$(x)_k = \frac{\Gamma (x + 1)}{\Gamma(x - k + 1)}.$$ So in terms of Gamma functions, the above sum can be rewritten as \begin{align*} S &= \frac{\Gamma (z_1 + 1)}{\Gamma (1 - z_2)} \sum_{k = 0}^\infty \frac{\Gamma (1 - z_2 - k)}{\Gamma (z_1 - k + 1)}\\ &= -\frac{\Gamma (z_1 + 1) \sin (z_1 \pi)}{\Gamma (1 - z_2) \sin(z_2 \pi)} \sum_{k = 0}^\infty \frac{\Gamma (k - z_1)}{\Gamma (k + z_2)} \tag1\\ &= -\frac{\Gamma (z_1 + 1) \sin(z_1 \pi)}{\Gamma (z_1 + z_2) \Gamma (1 - z_2) \sin(z_2 \pi)} \sum_{k = 0}^\infty \frac{\Gamma (k - z_1) \Gamma (z_1 + z_2)}{\Gamma (k + z_2)}\\ &= -\frac{\Gamma (z_1 + 1) \Gamma (z_2)}{\Gamma (z_1 + z_2) \Gamma (z_1) \Gamma (1 - z_1)} \sum_{k = 0}^\infty \text{B}(k - z_1, z_1 + z_2) \tag2\\ &= -\frac{\Gamma (z_1 + 1) \Gamma (z_2)}{\Gamma (z_1 + z_2) \Gamma (z_1) \Gamma (1 - z_1)} \sum_{k = 0}^\infty \int_0^1 t^{k - z_1 - 1} (1 - t)^{z_1 + z_2 - 1} \, dt \tag3\\ &= -\frac{\Gamma (z_1 + 1) \Gamma (z_2)}{\Gamma (z_1 + z_2) \Gamma (z_1) \Gamma (1 - z_1)} \int_0^1 t^{-z_1 - 1} (1 - t)^{z_1 + z_2 - 1} \sum_{k = 0}^\infty t^k \, dt \tag4\\ &= -\frac{\Gamma (z_1 + 1) \Gamma (z_2)}{\Gamma (z_1 + z_2) \Gamma (z_1) \Gamma (1 - z_1)} \int_0^1 t^{-z_1 - 1} (1 - t)^{z_1 + z_2 - 1} \cdot \frac{1}{1 - t} \, dt \tag5\\ &= -\frac{\Gamma (z_1 + 1) \Gamma (z_2)}{\Gamma (z_1 + z_2) \Gamma (z_1) \Gamma (1 - z_1)} \int_0^1 t^{-z_1 - 1} (1 - t)^{z_1 + z_2 - 2} \, dt\\ &= -\frac{\Gamma (z_1 + 1) \Gamma (z_2)}{\Gamma (z_1 + z_2) \Gamma (z_1) \Gamma (1 - z_1)} \text{B}(-z_1, z_1 + z_2 - 1) \tag6\\ &= -\frac{\Gamma (z_1 + 1) \Gamma (z_2)}{\Gamma (z_1 + z_2) \Gamma (z_1) \Gamma (1 - z_1)} \cdot \frac{\Gamma (-z_1) \Gamma (z_1 + z_2 - 1)}{\Gamma (z_2 - 1)} \tag7\\ &= \frac{z_2 - 1}{z_1 + z_2 - 1} \tag8 \end{align*}

Thus $$\sum_{k = 0}^\infty \frac{(z_1)_k}{(-z_2)_k} = \frac{z_2 - 1}{z_1 + z_2 - 1},$$ and is the simplified expression your Gamma function term reduces to.

Explanation

(1) Using Euler's reflection formula $\Gamma (x) \Gamma (1 - x) = \dfrac{\pi}{\sin (\pi x)}$.

(2) Using the property of the Beta function of $\text{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$ together with Euler's reflection formula.

(3) Using the integral representation for the Beta function of $\text{B}(x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt$.

(4) Interchanging the summation with the integration.

(5) Summing the series which is geometric.

(6) Identifying the integral as corresponding to the Beta function.

(7) As in the first part of (2).

(8) Making use of property for the Gamma function of $\Gamma (z + 1) = z \Gamma (z)$.

Comment on Convergence

For convergence in the Beta integral appearing in (2) we require $\text{Re}(z_1 + z_2) > 0$ and $\text{Re}(k - z_1) > 0$ implying $\text{Re}(z_1) < 0$ as $k = 0,1,2,\ldots$.

For convergence in the Beta integral appearing in (6) we require $\text{Re}(z_1 + z_2) > 1$ and $\text{Re}(z_1) < 0$.

So the series converges at least for $\text{Re}(z_1 + z_2) > 1$ and $\text{Re}(z_1) < 0$. As for values outside this range I cannot say.

Answer 3

Another approach is through the hypergeometric function.

In fact the ratio can be written in terms of Rising Factorials and then as terms of the hypergeometric sum $$ {{z_{\,1} ^{\,\underline {\,k\,} } } \over {\left( { - z_{\,2} } \right)^{\,\underline {\,k\,} } }} = {{\left( { - 1} \right)^{\,k} \left( { - z_{\,1} } \right)^{\,\overline {\,k\,} } } \over {\left( { - 1} \right)^{\,k} \left( {z_{\,2} } \right)^{\,\overline {\,k\,} } }} = {{\left( { - z_{\,1} } \right)^{\,\overline {\,k\,} } 1^{\,\overline {\,k\,} } } \over {\left( {z_{\,2} } \right)^{\,\overline {\,k\,} } }}{1 \over {k!}} $$

The infinite sum is therefore the hypergeometric function computed at the unitary value of its argument $$ \sum\limits_{k = 0}^\infty {{{z_{\,1} ^{\,\underline {\,k\,} } } \over {\left( { - z_{\,2} } \right)^{\,\underline {\,k\,} } }}} = {}_2F_{\,1} \left( {\left. {\matrix{ { - z_{\,1} ,1} \cr {z_{\,2} } \cr } \;} \right|\;1} \right) $$ which in virtue of the Gauss theorem gives $$ \eqalign{ & \sum\limits_{k = 0}^\infty {{{z_{\,1} ^{\,\underline {\,k\,} } } \over {\left( { - z_{\,2} } \right)^{\,\underline {\,k\,} } }}} = {}_2F_{\,1} \left( {\left. {\matrix{ { - z_{\,1} ,1} \cr {z_{\,2} } \cr } \;} \right|\;1} \right) = {{\Gamma (z_{\,2} )\Gamma (z_{\,2} + z_{\,1} - 1)} \over {\Gamma (z_{\,2} + z_{\,1} )\Gamma (z_{\,2} - 1)}} = \cr & = {{\left( {z_{\,2} - 1} \right)\Gamma (z_{\,2} - 1)\Gamma (z_{\,2} + z_{\,1} - 1)} \over {\left( {z_{\,2} + z_{\,1} - 1} \right)\Gamma (z_{\,2} + z_{\,1} )\Gamma (z_{\,2} - 1)}} = {{\left( {z_{\,2} - 1} \right)} \over {\left( {z_{\,2} + z_{\,1} - 1} \right)}}\quad \left| {\;1 < {\mathop{\rm Re}\nolimits} \left( {z_{\,1} } \right) + {\mathop{\rm Re}\nolimits} \left( {z_{\,2} } \right)} \right. \cr} $$ including the range of convergence.