Prove that $\sum_{k=0}^{m}\frac{\binom{m}{k}}{\binom{n}{k}}=\frac{n+1}{n+1-m}$ Prove that $\sum_{k=0}^{m}\dfrac{\binom{m}{k}}{\binom{n}{k}}=\dfrac{n+1}{n+1-m}$.
We know, n > m. From the right side. we have $\dfrac{n+1}{n+1-m}=\dfrac{1}{1-\dfrac{m}{n+1}}$. since n > m. $0<\dfrac{m}{n+1}<1$. Then
$\dfrac{1}{1-\dfrac{m}{n+1}}=1+\dfrac{m}{n+1}+(\dfrac{m}{n+1})^2+...$
I don't know what to do next? Did I choose the right path?

Partial Fractions
Using the Heaviside Method for Partial Fractions, we get
$$
\begin{align}
\sum_{k=0}^m\frac{\binom{m}{k}}{\binom{n}{k}}
&=1+\sum_{k=1}^m\frac{m(m-1)(m-2)\dots(m-k+1)}{n(n-1)(n-2)\dots(n-k+1)}\tag{1}\\
&=1+m\sum_{k=1}^m\sum_{j=0}^{k-1}\frac{(-1)^{k-j-1}\binom{m-1}{k-1}\binom{k-1}{j}}{n-j}\tag{2}\\
&=1+m\sum_{k=1}^m\sum_{j=0}^{k-1}\frac{(-1)^{k-j-1}\binom{m-1}{j}\binom{m-j-1}{k-j-1}}{n-j}\tag{3}\\
&=1+m\sum_{j=0}^{m-1}\sum_{k=j+1}^m\frac{(-1)^{k-j-1}\binom{m-1}{j}\binom{m-j-1}{k-j-1}}{n-j}\tag{4}\\[6pt]
&=1+\frac{m}{n-m+1}\tag{5}\\[6pt]
&=\frac{n+1}{n-m+1}\tag{6}
\end{align}
$$
Explanation:
$(1)$: break out the $k=0$ term and expand numerator and denominator
$(2)$: apply the Heaviside Method to the fraction in the sum
$(3)$: $\binom{m-1}{k-1}\binom{k-1}{j}=\binom{m-1}{j}\binom{m-j-1}{k-j-1}$
$(4)$: switch the order of summation
$(5)$: $\sum\limits_{k=j+1}^m(-1)^{k-j-1}\binom{m-j-1}{k-j-1}=0^{m-j-1}=\big[j=m-1\big]$
$(6)$: addition

Hockey-Stick Identity
$$
\begin{align}
\sum_{k=0}^m\frac{\binom{m}{k}}{\binom{n}{k}}
&=\sum_{k=0}^m\frac{\frac{m!}{(m-k)!}}{\frac{n!}{(n-k)!}}\tag7\\
&=\frac{m!(n-m)!}{n!}\sum_{k=0}^m\frac{(n-k)!}{(m-k)!(n-m)!}\tag8\\
&=\frac1{\binom{n}{m}}\sum_{k=0}^m\binom{n-k}{n-m}\tag9\\
&=\frac1{\binom{n}{m}}\binom{n+1}{n-m+1}\tag{10}\\[3pt]
&=\frac{n+1}{n-m+1}\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(7)$: expand the binomial coefficients as factorials and cancel
$\phantom{1}(8)$: shuffle factors and throw in $\frac{(n-m)!}{(n-m)!}$
$\phantom{1}(9)$: recognize binomial coefficients
$(10)$: Hockey-stick identity
$(11)$: simplify
