Answer

Note that the process till the end is a combination of multiple binomial process, such that any process lasts till the first success. Suppose we sign a paper and keep this in the drawer. Now the total signed papers in the drawer is \(k\) out of \(n\) and the \(P(\)success\()\) = \(\frac{n-k}{n}\) and \(E[\)draws till next unsigned paper\(] = \frac{1}{p} = \frac{n}{n-k}\). Total draws

\[ \begin{aligned} E &= \frac{n}{1} + \frac{n}{2} + \cdots + \frac{n}{n}\newline &= n(1 + \frac{1}{2} + \cdots + \frac{1}{n})\newline \lim_{n \to large} E &= n \log(n) \end{aligned} \]