Closed-form formula to evaluate $\sum_{k = 0}^{m} \binom{2m-k}{m}\cdot 2^k$

Question

Inspired by this question I'm trying to prove that $$\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!} \approx e^{\frac{x}{2}}$$ So I needed to find the value of $$\frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{e^{\frac{x}{2}}} = \frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{\frac{x}{2}^k}{k!}} \\ = \lim_{m \to \infty} \frac{\sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{\sum_{k = 0}^{m} \frac{\frac{x}{2}^k}{k!}} = \lim_{m \to \infty} \sum_{k = 0}^{m} \frac{\frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{\frac{\frac{x}{2}^k}{k!}} = \lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}\cdot\frac{k!}{\frac{x}{2}^k} \\ = \lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\cdot 2^k $$ Now, since $$\sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}\cdot 2^k=\sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!} \frac{m!}{m!}\cdot 2^k=\sum_{k = 0}^{m} \binom{m}{2m} \cdot \binom{2m-k}{m}\cdot 2^k\\=\binom{m}{2m} \cdot \sum_{k = 0}^{m} \binom{2m-k}{m}\cdot 2^k=\frac{\sum_{k = 0}^{m} \binom{2m-k}{m}\cdot 2^k}{\binom{2m}{m}}$$ I only need to prove that $$\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!} = \lim_{m \to \infty} \frac{\sum_{k = 0}^{m} \binom{2m-k}{m}\cdot 2^k}{\binom{2m}{m}} \approx e^{\frac{x}{2}}$$ so the question: is there a closed-form formula to evaluate $\sum_{k = 0}^{m} \binom{2m-k}{m}\cdot 2^k$ (Bessel functions are not allowed)?

Answer 1

The comment by @Byron Schmuland has really helped me out. Since $$\sum_{k=0}^{n}\dfrac{\binom{2n-k}{n}}{2^{2n-k}}=1$$ one can write $$\sum_{k=0}^{n}\dfrac{\binom{2n-k}{n}}{2^{2n-k}}=\sum_{k=0}^{n}\frac{\binom{2n-k}{n}}{\frac{2^{2n}}{2^k}}=\sum_{k=0}^{n}\frac{\binom{2n-k}{n} \cdot 2^k}{2^{2n}}=\frac{1}{4^n} \cdot \sum_{k=0}^{n}\binom{2n-k}{n} \cdot 2^k = 1\\ \Rightarrow \sum_{k = 0}^{n} \binom{2n-k}{n}\cdot 2^k=4^n$$ thus $$\lim_{m \to \infty} \frac{\sum_{k = 0}^{m} \binom{2m-k}{m}\cdot 2^k}{\binom{2m}{m}} = \lim_{m \to \infty} \frac{4^m}{\binom{2m}{m}} = \lim_{m \to \infty} \frac{4^m \cdot (m!)^2}{(2m)!}$$ Now by applying Stirling's approximation we get $$\lim_{m \to \infty} \frac{4^m \cdot (m!)^2}{(2m)!} = \lim_{m \to \infty} \frac{4^m \cdot \left(\sqrt{2\pi m} \cdot \left(\frac{m}{e}\right)^m\right)^2}{\sqrt{4\pi m} \cdot \left(\frac{2m}{e}\right)^{2m}} = \lim_{m \to \infty} 4^m \frac{2\pi m}{2 \sqrt{\pi m}} \frac{\left(\frac{m}{e}\right)^{2m}}{\left(\frac{2m}{e}\right)^{2m}}\\= \lim_{m \to \infty} \sqrt{\pi m} \cdot 4^m \cdot \frac{m^{2m}}{(2m)^{2m}} = \lim_{m \to \infty} \sqrt{\pi m} \cdot 4^m \cdot \frac{m^{2m}}{m^{2m} \cdot 4^m} = \lim_{m \to \infty} \sqrt{\pi m} = +\infty$$ So, recombining the pieces, we finally get $$\frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{e^{\frac{x}{2}}} = \lim_{m \to \infty} \sqrt{\pi m}\Rightarrow\\ \lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!} \approx \sqrt{\pi m} \cdot e^{\frac{x}{2}}$$