Summing something that looks binomial

Question

I have a sum that looks like

$\sum_{j=0}^{k/2} \frac{k!}{(k-2j)!j!(2d)^j }p^{k-j}(1-p)^j$.

$p \in (0,1), d$ is an integer $> 1$

I am wondering is at least an approximate sum is known.

Answer 1

I'm replacing $k = 2r$ because I don't want to deal with the issue of odd $k$.

The exact sum is known to be expressible in terms of Tricomi's form of confluent hypergeometric function $U(a,b,x)$. When $$ U(a,b,x) = \frac1{\Gamma(a)}\int_0^\infty t^{a-1}(1-t)^{b-a-1}e^{-xt}\,dt $$ (sometimes called CFGs of the second kind). In terms of that, the formal sum is

$$(-1)^r\left( \frac{(4p(1-p)}{(2d)!} \right)^r U\left((-r,\frac12,-\frac{(2d)!p}{4(1-p)}\right) $$

The asymptotic behavior of this CFG looks like $$ C(-1)^r \left(\frac{(2d)!p}{4(1-p)}\right)^r $$ where $C$ is a generalized hypergeometric series with $1$ as its leading term.

The $(-1)^r$ factor will disappear against the identical factor inside the behavior of the CFG, leaving an asymptotic behavior of $$ \left(\frac{(2d)!p}{4(1-p)}\right)^r $$

And this almost exactly cancels the factor $\left( \frac{(4p(1-p)}{(2d)!} \right)^r$ outside the CFG, leaving a result of $p^r$.

So by this reasoning this sum should go like $p^{k/2}$. However, this all takes lots of liberties with convergence of series and so forth, so I would not trust it too heavily.