Equation for Limit of Infinite Series

Question

Is it possible to represent this $$\lim_{k\to \infty} \left(\sum_{n = 0}^\infty \frac{x^{2n}(k-n)!}{(k+n)!}\right)$$ as a single equation in terms of $x$? If so, what would that equation be? Thank you!

Answer 1

Hint

Recall that $\lim_{k \to \infty} \frac{(k-n)!}{(k+n)!} = 0$ for all $n>0$

Answer 2

$$\sum_{n=0}^{k}\frac{x^{2n}(k-n)!}{(k+n)!} = 1+\sum_{n=1}^{k}\frac{x^{2n}\,\Gamma(k+1-n)}{\Gamma(k+1+n)}=1+\sum_{n=1}^{k}\frac{x^{2n}}{\Gamma(2n)}\,B(2n,k+1-n)$$ equals: $$ 1+\int_{0}^{1}\sum_{n=1}^{k}\frac{x^{2n}}{(2n-1)!}\,y^{2n-1}(1-y)^{k-n}\,dy \leq 1+\int_{0}^{1}(1-y)^k\frac{x}{\sqrt{1-y}}\sinh\left(\frac{xy}{\sqrt{1-y}}\right)\,dy$$ but for any fixed $x$ the function $(1-y)^k\frac{x}{\sqrt{1-y}}\sinh\left(\frac{xy}{\sqrt{1-y}}\right)$ is uniformly convergent to $0$ on the interval $(0,1)$ for $k\to +\infty$. It follows that the given limit is pointwise $1$ for any $x>0$.