Calculating expection of number of draws necessary to capture $m$ marked animals I'm working on the following problem, but I'm stuck and I hope anyone can provide some help:

A population of $N$ animals has had a certain number $a$ of its members captured, marked and then released. Show that the probability $p_n$ that it is necessary to capture $n$ animals in order to obtain $m$ which have been marked is $$p_n=  \frac{a}{N}\binom{a-1}{m-1}\binom{N-a}{n-m}\left/\binom{N-1}{n-1}\right.,$$ where $m \leq n \leq N-a+m$. Hence, show that $$\frac{a}{N}\binom{a-1}{m-1}\frac{(N-a)!}{(N-1)!}\sum_{n=m}^{N-a+m}\frac{(n-1)!(N-n)!}{(n-m)!(N-a+m-n)!}=1,$$ and that the expectation of $n$ is $\frac{N+1}{a+1}m$. 

Well, I was able to do every step, but I'm stuck at what to do to show that $\mathbb{E}(n)=\frac{N+1}{a+1}m$. Any help would be appreciated!

Let $X$ be a random variable representing the number of animals it is necessary to capture in order to obtain $m$ marked animals, from a total population of $N$ animals, $a$ of which are marked. Then,
$$p_n = \mathbb{P}(X=n) = \frac{a}{N}\binom{a-1}{m-1}\binom{N-a}{n-m}\left/\binom{N-1}{n-1}\right.$$
for $n = m,m+1,\ldots, N-a+m$. The task is to find $\mathbb{E}(X)$.
By definition, $$\mathbb{E}(X) = \sum_{n=m}^{N-a+m} np_n$$
Let $X'=X-m$, $x = n-m$ and $p_x = \mathbb{P}(X'=x)$ for $x=0,1,\ldots,N-a$.
Then $p_x = p_n$ and
\begin{align}
\mathbb{E}(X) &= \sum_{x=0}^{N-a} (x+m)p_x \\
&= m + \sum_{x=0}^{N-a} xp_x \\
&= m + \sum_{x=1}^{N-a} x \frac{a}{N}\binom{a-1}{m-1}\binom{N-a}{x}\left/\binom{N-1}{x+m-1}\right. \\
&= m + \frac{a}{N}\binom{a-1}{m-1} \sum_{x=1}^{N-a} x \binom{N-a}{x}\left/\binom{N-1}{x+m-1}\right. \\
&= m + \frac{m}{N}\binom{a}{m} \sum_{x=1}^{N-a} (N-a)\binom{N-a-1}{x-1}\left/\binom{N-1}{x+m-1}\right. \\
&= m + \frac{m(N-a)}{a+1} \sum_{x=1}^{N-a} \frac{a+1}{N} \binom{a}{m}\binom{N-a-1}{x-1}\left/\binom{N-1}{x+m-1}\right. \\
&= m + \frac{m(N-a)}{a+1} \sum_{n=m+1}^{N-(a+1)+(m+1)} \frac{a+1}{N} \binom{(a+1)-1}{(m+1)-1}\binom{N-(a+1)}{n-(m+1)}\left/\binom{N-1}{n-1}\right.
\end{align}
Let $Y$ be a random variable representing the number of animals it is necessary to capture in order to obtain $m+1$ marked animals, from a total population of $N$ animals, $a+1$ of which are marked. Then,
$$\mathbb{P}(Y=n) = \frac{a+1}{N} \binom{(a+1)-1}{(m+1)-1}\binom{N-(a+1)}{n-(m+1)}\left/\binom{N-1}{n-1}\right.$$
for $n = m+1,m+2,\ldots,N-(a+1)+(m+1)$. Therefore,
\begin{align}
\mathbb{E}(X) &= m + \frac{m(N-a)}{a+1} \sum_{n=m+1}^{N-(a+1)+(m+1)} \mathbb{P}(Y=n) \\
&= m + \frac{m(N-a)}{a+1} \\
&= \frac{N+1}{a+1}m
\end{align}
as required.
