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!
 A: For the  probabilities when  we multiply them  recursively we  get two
shuffled         contributions         $a^{\underline{m}}$         and
$(N-a)^{\underline{n-m}}$ in the  numerator and $N^{\underline{n}}$ in
the denominator.  The last item  must be a  marked animal and  for the
animals which came before that we may choose their positions according
to ${n-1\choose m-1}.$ We thus get for the probabilities
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{P}[X=n] = \frac{a^{\underline{m}} (N-a)^{\underline{n-m}}}
{N^{\underline{n}}} {n-1\choose m-1}.}$$
Working to convert this into binomial coefficients we get
$$\mathrm{P}[X=n] =
m! \times {a\choose m} \times (n-m)! \times {N-a\choose n-m}
\times {N\choose n}^{-1} \frac{1}{n!} \times {n-1\choose m-1}
\\ = m! \times {a\choose m} \times (n-m)! \times {N-a\choose n-m}
\times {N\choose n}^{-1} \frac{1}{n!} \times \frac{m}{n} {n\choose m}
\\ = \frac{m}{n} {a\choose m} {N-a\choose n-m}
{N\choose n}^{-1}.$$
To verify that  this is a probability distribution we  observe that we
must draw  at least $m$ animals  (initial sequence all marked)  and at
most $N-a+m$ animals (non-marked individuals come first).
We get for the probabilities
$$m {a\choose m}
\sum_{n=m}^{N-a+m} {N-a\choose n-m}
\frac{1}{n} {N\choose n}^{-1}
= m {a\choose m}
\sum_{n=0}^{N-a} {N-a\choose n}
\frac{1}{n+m} {N\choose n+m}^{-1}
\\ = m {a\choose m}
\sum_{n=0}^{N-a} {N-a\choose n}
\mathrm{B}(n+m, N-n-m+1)
\\ = m {a\choose m}
\sum_{n=0}^{N-a} {N-a\choose n}
\int_0^1 t^{n+m-1} (1-t)^{N-n-m} \; dt
\\ = m {a\choose m}
\int_0^1 t^{m-1} (1-t)^{N-m}
\sum_{n=0}^{N-a} {N-a\choose n} \frac{t^n}{(1-t)^n}
\; dt
\\ = m {a\choose m}
\int_0^1 t^{m-1} (1-t)^{N-m}
\left(1+\frac{t}{1-t}\right)^{N-a}
\; dt
\\ = m {a\choose m}
\int_0^1 t^{m-1} (1-t)^{a-m}
\; dt
= m {a\choose m} \mathrm{B}(m, a-m+1)
\\ = m {a\choose m} \frac{1}{m} {a\choose m}^{-1} = 1.$$
Now for the expectation we re-use this same computation, getting
$$m + m {a\choose m}
\sum_{n=0}^{N-a} {N-a\choose n} n
\int_0^1 t^{n+m-1} (1-t)^{N-n-m} \; dt
\\ = m + m {a\choose m}
\int_0^1 t^{m-1} (1-t)^{N-m}
\sum_{n=1}^{N-a} {N-a\choose n} n
\frac{t^n}{(1-t)^n} \; dt
\\ = m + (N-a) m {a\choose m}
\int_0^1 t^{m-1} (1-t)^{N-m}
\sum_{n=1}^{N-a} {N-a-1\choose n-1}
\frac{t^n}{(1-t)^n} \; dt
\\ = m + (N-a) m {a\choose m}
\int_0^1 t^{m} (1-t)^{N-m-1}
\sum_{n=1}^{N-a} {N-a-1\choose n-1}
\frac{t^{n-1}}{(1-t)^{n-1}} \; dt
\\ = m + (N-a) m {a\choose m}
\int_0^1 t^{m} (1-t)^{N-m-1}
\left(1+\frac{t}{1-t}\right)^{N-a-1}
\; dt
\\ = m + (N-a) m {a\choose m}
\int_0^1 t^{m} (1-t)^{a-m}\; dt
\\ = m + (N-a) m {a\choose m}
\mathrm{B}(m+1, a-m+1)
\\ = m + (N-a) m {a\choose m} \frac{1}{m+1} {a+1\choose m+1}^{-1}
= m + (N-a) m \frac{1}{a+1}.$$
This indeed yields
$$\bbox[5px,border:2px solid #00A000]{
\mathrm{E}[X] = m \frac{N+1}{a+1}}$$
as claimed. As a sanity check when all animals have been marked we get
$m$ marked animals after drawing $m$ individuals no matter what, which
is the correct result.
A: 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.
