Find closed form of sum of fraction of binomial coefficients can somebody give me a hint for this exercise, where I have to find the specific closed form?

$\sum_{k=0}^m \frac{\binom{m}{k}}{\binom{n}{k}}, m,n\in\mathbb{N}$ and $m\leq n$

What I have done so far:
$\sum_{k=0}^m \frac{\binom{m}{k}}{\binom{n}{k}} = \sum_{k=0}^m \frac{\frac{m!}{(m-k)!\cdot k!}}{\frac{n!}{(n-k)!\cdot k!}} = \sum_{k=0}^m  \frac{m!}{n!}\cdot \frac{(n-k)!}{(m-k)!} = \frac{m!}{n!}\cdot \sum_{k=0}^m  \frac{(n-k)!}{(m-k)!}$ 
Look at example 1: $n=8, m=5$
$\frac{5!}{8!}\cdot(\frac{8!}{5!} + \frac{7!}{4!} +\frac{6!}{3!} +\frac{5!}{2!} + \frac{4!}{1!} +\frac{3!}{0!}) = \\\frac{5!}{8!} \cdot (8\cdot7\cdot6+7\cdot6\cdot5+6\cdot5\cdot4+5\cdot4\cdot3+4\cdot3\cdot2+3\cdot2\cdot1) = \\
\frac{5!}{8!} \cdot (336+210+120+60+24+6) = \frac{5!}{8!}\cdot 756 = 2.25$
I can't find a pattern.
Edit 1: Maybe there is an approach with recursion.
Edit 2: Okay I found the solution. 
$\sum_{k=0}^m \frac{m!}{n!}\cdot \sum_{k=0}^m  \frac{(n-k)!}{(m-k)!}= \frac{m!}{n!}\cdot\frac{1}{4}\cdot t\cdot(t+1)\cdot(t+2)\cdot(t+3)$ with $t=(n-m)!$
Edit 3: This formula works well for example 1, but fails for example 2: $n=9, m=3$
$\frac{3!}{9!}\cdot(\frac{9!}{3!} + \frac{8!}{2!} +\frac{7!}{1!} +\frac{6!}{0!})  = \\\frac{3!}{9!} \cdot (9\cdot8\cdot7\cdot6\cdot5\cdot4+8\cdot7\cdot6\cdot5\cdot4\cdot3+7\cdot6\cdot5\cdot4\cdot3\cdot2+6\cdot5\cdot4\cdot3\cdot2\cdot1) = \\
\frac{3!}{9!} \cdot (60480+20160+5040+720) = \frac{3!}{9!}\cdot 86400 = 1.428$
So I have still no general solution. Can someone help?
 A: With $m\le n$ we have the sum
$$\sum_{k=0}^m {m\choose k} {n\choose k}^{-1}
= \sum_{k=0}^m \frac{m!}{(m-k)! k!} \frac{k! (n-k)!}{n!}
\\ = \frac{m!}{n!} \sum_{k=0}^m \frac{(n-k)!}{(m-k)!}
= {n\choose m}^{-1} \sum_{k=0}^m {n-k\choose m-k}
\\ = {n\choose m}^{-1} \sum_{k=0}^m [z^{m-k}] (1+z)^{n-k}
= {n\choose m}^{-1} [z^m] (1+z)^n \sum_{k=0}^m z^k (1+z)^{-k}.$$
We may extend $k$ beyond $m$ due to the $z^k$ term and the coefficient
extractor $[z^m]$ in front, getting
$${n\choose m}^{-1} [z^m] (1+z)^n \sum_{k\ge 0} z^k (1+z)^{-k}
= {n\choose m}^{-1} [z^m] (1+z)^n \frac{1}{1-z/(1+z)}
\\ = {n\choose m}^{-1} [z^m] (1+z)^{n+1}   
= {n\choose m}^{-1} {n+1\choose m}
= \frac{n+1}{n+1-m}.$$
A: Let 
$$S(m,n):=\sum_{k=0}^m \frac{\binom{m}{k}}{\binom{n}{k}}$$
We are going to prove:
$$S(m,n)=\frac{n+1}{n+1-m}.\tag{1}$$
Obviously (1) is valid for $m=0$ and arbitray $n\ge 0$. Further, if (1) is valid for $(m-1,n-1)$ it is valid for $(m,n)$ as well:
$$
S(m,n):=1+\sum_{k=1}^m \frac{\frac mk\binom{m-1}{k-1}}{\frac nk\binom{n-1}{k-1}}
=1+\frac{m}{n} S(m-1,n-1)\\
\stackrel{I.H.}{=}1+\frac{m}{n}\frac{n}{n+1-m}=\frac{n+1}{n+1-m}.
$$
Thus by induction (1) is valid for arbitrary $0\le m \le n$.
A: Hint:
Use the beta function
\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1(1-x)^kx^{n-k}\,dx
\end{align*}
