= Solution
Use standard neutral coalescent time units, so each pair of ancestral lineages merges at rate 1. With $j$ lineages, there are $\binom j2$ possible pairs; consequently the successive waiting times are independent with
$$
T_j\sim\operatorname{Exp}(\lambda_j),\qquad \lambda_j=\frac{j(j-1)}2.
$$
During this epoch there are $j$ branches, each of length $T_j$. Hence the <total branch length of a neutral coalescent> is $L=\sum_{j=2}^n jT_j$, and <linearity of expectation> gives
$$
\boxed{EL=\sum_{j=2}^n\frac{j}{\lambda_j}=2\sum_{i=1}^{n-1}\frac1i.}
$$
The root branch above the sample's common ancestor is excluded: its mutations would be shared by every chromosome and would not be <segregating sites> in the sample.
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