Falling and rising factorial series identity

Question

$$ \sum_{n=0}^\infty\prod_{m=1}^n\frac{x-m+1}{km} = \sum_{n=0}^\infty\prod_{m=1}^n\frac{x+m-1}{(k+1)m} $$

I noticed this identity that relates the falling and rising factorials using a power series. How can I derive this identity?

For k>=1, both expressions converge to $$ \frac{(k+1)^x}{k^x} $$If k>1, it seems to work for real x, but if k=1, only for x>=0.

We can use the product $ \prod_{m=0}^n{x+m} $ to build a triangle similar to Pascal's triangle, given by a recurrence relation $$ s(n,k)=ks(n-1,k-1)+s(n,k-1) $$, which looks like $$ k\begin{array} {|r|r|}\hline 1 & 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 1 & 0 & 0 & 0 & 0 \\ \hline 1 & 3 & 2 & 0 & 0 & 0 \\ \hline 1 & 6 & 11 & 6 & 0 & 0 \\ \hline 1 & 10 & 35 & 50 & 24 & 0 \\ \hline 1 & 15 & 85 & 225 & 274 & 120 \\ \hline \end{array} $$ If I knew how to parameterize this triangle, it might be easier. The second column contains the triangle numbers, and the third column, courtesy of OEIS, is apparently $ \sum_{k=0}^{n-1}\frac{k(k+1)^2}{2}. $ I'm not sure how to get the other columns.

Answer 1

For $|k| > 1$, the left-hand side (LHS) of the proposed identity reduces to

\begin{align*} \text{LHS} &= \sum_{n=0}^{\infty} \binom{x}{n} \frac{1}{k^n} = \left(1+\frac{1}{k}\right)^x = \left( \frac{k+1}{k} \right)^x \end{align*}

where we utilized the binomial series in the second step. On the other hand, for $|k+1| > 1$ the right-hand side (RHS) becomes

\begin{align*} \text{RHS} &= \sum_{n=0}^{\infty} \binom{-x}{n} \left(-\frac{1}{k+1}\right)^n = \left(1-\frac{1}{k+1}\right)^{-x} = \left( \frac{k+1}{k} \right)^x. \end{align*}

Therefore they are equal.

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Addendum. Below is the relevant result regarding the binomial series:

Theorem. Let $x$ be any complex number. Then

$$ (1 + z)^{x} = \sum_{n=0}^{\infty} \binom{x}{n} z^n $$

holds (at least) for $|z| < 1$. Here, the left-hand side is defined as $e^{x \log(1+z)}$, where $\log$ is the principal branch cut of the complex logarithm.