= Solution
First $0\le M_n\le\sum_iE_i$, so <linearity of expectation> gives $\mathbb E M_n\le n<\infty$. The <tail integral formula for expectation>, followed by the finite <binomial theorem> expansion, gives
$$
\begin{aligned}
\mathbb E M_n&=\int_0^\infty\left[1-(1-e^{-x})^n\right]dx\\
&=\sum_{k=1}^n(-1)^{k+1}\binom nk\int_0^\infty e^{-kx}dx\\
&=\boxed{\sum_{k=1}^n\binom nk\frac{(-1)^{k+1}}k.}
\end{aligned}
$$
The exchange involves only a finite sum. As a useful check, substitution $t=e^{-x}$ rewrites the integral as $\int_0^1[1-(1-t)^n]/t\,dt=\sum_{j=0}^{n-1}\int_0^1(1-t)^jdt=\sum_{k=1}^n1/k$, the <harmonic number> $H_n$.
Back to article page
