Simplifying:∑n=1∞nan(n−m)!(n+z)(x−c)n−m

Question

Haiden Frazier · Accepted Answer

It may be expressed as a general hypergeometric:

\sum_{n = 1}^{\infty} \frac{n a^{n} {(x - c)}^{n - m}}{(n - m)! (n + z)} = \frac{a_{2} F_{2} (2, 1 + z; 2 - m, 2 + z; a (x - c))}{(1 - m)! (1 + z) {(x - c)}^{m - 1}}

It is simply the definition of the hypergeometric series. Use notation

x^{(n)}

for the "rising factorial"

x^{(n)} = \prod_{j = 0}^{n - 1} (x + j)

2 F_{2} (2, 1 + z; 2 - m, 2 + z; a (x - c)) = \sum_{k = 0}^{\infty} \frac{2^{(k)} {(1 + z)}^{(k)}}{{(2 - m)}^{(k)} {(2 + z)}^{(k)} k!} {(a (x - c))}^{k}

But note

\frac{2^{(k)}}{k} \neq (k + 1)

and

\frac{{(1 + z)}^{(k)}}{{(2 + z)}^{(k)}} = \frac{1 + z}{1 + k + z}

, so

\frac{a_{2} F_{2} (2, 1 + z; 2 - m, 2 + z; a (x - c))}{(1 - m)! (1 + z) {(x - c)}^{m - 1}} = \sum_{n = 1}^{\infty} \frac{n}{(n - m)! (n + z)} a^{n} {(x - c)}^{n - m}

Simplifying: \sum_{n=1}^\infty\frac{na^n}{(n-m)!(n+z)}(x-c)^{n-m}