# Evaluation of $\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}$

I have problems evaluating the following limit:

$$\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}$$

What causes problems in particular is that I am unsure how to behave when there is a sum which becomes a series. I am aware that I need to evaluate all limits at the same time. So, I am not sure if I can use any knowledge about series (e.g. radius of convergence, which should give me the ratio of the coefficents of the series so that I can probably write down the limit) to get more information because it will only be a series when taking the limit. Using Sterling is in question was not that helpful either, because I didn't figure out any way of properly arranging the terms.

I need to show that it is only $\approx \exp(x/2)$ which would be enough.

Any help would be appreciated!

## 2 Answers

$$\frac{1}{(2m)!}\sum_{k=0}^{m}\binom{m}{k} x^k (2m-k)!=\frac{1}{(2m)!}\int_{0}^{+\infty}\sum_{k=0}^{m}\binom{m}{k} x^k z^{2m-k} e^{-z}\,dz$$ and the RHS can be written as: $$\frac{1}{(2m)!}\int_{0}^{+\infty} \left(z(x+z)\right)^m e^{-z}\,dz = \frac{e^{x/2}}{(2m)!}\int_{x/2}^{+\infty} \left(z^2-\frac{x^2}{4}\right)^m e^{-z}\,dz$$ whose exact value is:

$$e^{x/2}\cdot \frac{x^{m+\frac{1}{2}}\cdot K_{m+\frac{1}{2}}\left(\frac{x}{2}\right)}{\sqrt{\pi}\cdot(m+1)!}$$ with $K$ being a modified Bessel function of the second kind.

I've posted a question about this question and I found that $$\frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{e^{\frac{x}{2}}} = \lim_{m \to \infty} \sqrt{\pi m}\Rightarrow\\ \lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!} \approx \sqrt{\pi m} \cdot e^{\frac{x}{2}}$$ without the use of Bessel functions.