0
$\begingroup$

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!

  • 0
    Please use [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). – 2017-02-12
  • 0
    How are you defining $(k-n)!$ for $n>k$? – 2017-02-12
  • 0
    @Clayton not very consequential. Just bring limit inside the sum. – 2017-02-12

2 Answers 2

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 $x>0$.

0

Hint

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

  • 2
    Can you justify taking the limit inside of this sum? It's not always obvious, eg. $\lim_{k \to \infty} 1/(k + n^2/k ) = 0$ but $\lim_{k \to \infty} \sum_{n=1}^\infty 1/(k +n^2/k) = \pi/2$ – 2017-02-12
  • 0
    @RRL I'll test uniform convergence by M test when I have time in a couple hours. – 2017-02-12