Graduate Is the following sum a part of any known generalized function?

[tworitdash]

I have a sum that looks like the following:

$\sum_{k = 0}^{\infty} \left( \frac{A}{A + k} \right)^{\eta} \frac{z^k}{k!}$

Here, A is positive real.

If \eta is an integer, this can be written as:

$\sum_{k = 0}^{\infty} \left( \frac{A(A +1)(A+2) \cdots (A + k - 1)}{(A + 1)(A+2)(A+3) \cdots (A + k)} \right)^{\eta} \frac{z^k}{k!}$

This is known to be a generalized hypergeometric function with \eta number of argument of type 1 and type 2 as well.

$\sum_{k = 0}^{\infty} \left( \frac{A(A +1)(A+2) \cdots (A + k - 1)}{(A + 1)(A+2)(A+3) \cdots (A + k)} \right)^{\eta} \frac{z^k}{k!} = _{\eta}F_{\eta} \left( A, A, A, ..., A; A+1, A+1, A+1, ... , A+1; z \right)$

However, this is possible because \eta is an integer. Can I approximate it to a nice form when it is not an integer? Or, can we determine where this infinite sum converges with some techniques?

 

[pasmith]

The radius of convergence is given by <br /> \lim_{k \to \infty} \left| \frac{\left(\frac{A}{A + k}\right)^\eta\frac{1}{k!}}{\left(\frac{A}{A + k + 1}\right)^\eta\frac{1}{(k+1)!}} \right| = \lim_{k \to \infty} \left|\left(1 + \frac{1}{A + k}\right)^{\eta}(k+1)\right| = \infty.

 

[tworitdash]

The radius of convergence is given by \lim_{k \to \infty} \left| \frac{\left(\frac{A}{A + k}\right)^\eta\frac{1}{k!}}{\left(\frac{A}{A + k + 1}\right)^\eta\frac{1}{(k+1)!}} \right| = \lim_{k \to \infty} \left|\left(1 + \frac{1}{A + k}\right)^{\eta}(k+1)\right| = \infty.

So, it is an absolutely converging function. That I get it as well. Is there a possibility to get an asymptotic value of this sum for non-integer \eta, as a function of \eta, and A?