Sum of products in an expected value A box contains $n$ balls numbered from 1 to $n$. Suppose you take a ball at a time, putting it back on the box, until you pick a ball twice. How many balls are you expected to take from the box?
Let $X$ be the r. v. of interest. Its support is every natural number from $2$ to $n+1$. If $k$ is one such number, to get the first repetition on the $k$-th pick, the $k-1$ previous ones must be distinct and $k$-th equal to some of those, therefore,
$$\begin{align}
\forall k\in\mathbb N\cap[2,\,n\!+\!1]\qquad \mathbb P(X=k) &=
\frac nn\frac {n-1}n\ldots\frac {n-(k-2)}n \;\cdot\; \frac{k-1}n =\\
&=\frac{(n-1)!}{(n-k+1)!}\frac{k-1}{n^{k-1}} =\\
&= \frac{k-1}n\,\prod_{l=0}^{k-2}\left(1-\frac{l}n\right)\quad,
\end{align}$$
so that
$$\begin{align}
E(X) &= \sum_{k=2}^{n+1}\,k\cdot\frac{k-1}n\,\prod_{l=0}^{k-2}\left(1-\frac{l}n\right) =\\
&=\frac1n\sum_{k=2}^{n+1}\,k(k-1)\prod_{l=0}^{k-2}\left(1-\frac{l}n\right)\quad.
\end{align}$$
However, the book answer is
$$E(X) = 2 + \sum_{k=1}^{n-1}\prod_{l=1}^k\left(1-\frac ln\right)\quad.$$
What am I doing wrong? Or, if the answers are actually the same, how to show that?
 A: Both answers are the same.  The solution in the question is clearly presented and holds up.  For reference, the answers it gives for $n=2, 3, \ldots, 7$ are
$$E(X) = 2,\frac{5}{2},\frac{26}{9},\frac{103}{32},\frac{2194}{625},\frac{1
   223}{324},\frac{472730}{117649}, \ldots$$
A simpler way to obtain the expectation is to sum the survival function $S_n(k) = \mathbb{P}(X\gt k), k\ge 0.$  This is the probability that the first $k$ balls will be unique:
$$S_n(k) = \prod_{l=1}^{k-1} \left(1 - \frac{l}{n}\right).$$
Clearly $S_n(0)=S_n(1)=1$ (the value, by definition, of an empty product). Accounting for their sum separately yields
$$E(X) = S_n(0) + S_n(1) + \sum_{j=2}^n S_n(j) = 2 + \sum_{k=1}^{n-1} S_n(k+1).$$
(writing $j=k+1$ for the last step).  That is the textbook answer and it gives the same sequence of values.  Incidentally, a useful closed form expression for these values is
$$E(X) = 1 + n \int_0^{\infty } e^{-n t} (1+t)^{n-1} \, dt.$$
For instance, expanding the log of the integrand through second order at $0$ and integrating that MacLaurin series gives the approximation $$E(X)\approx \frac{2}{3} + \sqrt{\frac{\pi n}{2}}$$ for large $n$; it turns out to have two significant figures of accuracy even for $n\ge 5$ and three for $n \ge 75.$
