= Solution
Use standard diploid coalescent time units of $2N_e$ generations, where $N_e$ is the fixed <effective population size>, and take $n\ge2$. In the neutral <Kingman's coalescent>, each unordered pair of ancestral lineages merges at rate one. With $j$ lineages, the next coalescence time therefore has an <exponential distribution> with rate
$$
\lambda_j=\binom j2=\frac{j(j-1)}2,\qquad E[T_j]=\frac{2}{j(j-1)}.
$$
The memoryless construction and the <Markov property> make these epoch durations independent. During epoch $j$ there are $j$ branches, each of duration $T_j$, so the <total branch length of a neutral coalescent> is
$$
L=\sum_{j=2}^n jT_j.
$$
By linearity of <expectation>,
$$
\boxed{E[L]=\sum_{j=2}^n\frac{2}{j-1}=2\sum_{i=1}^{n-1}\frac1i=2a_n,\qquad a_n=\sum_{i=1}^{n-1}\frac1i.}
$$
The ancestral branch above the sample's most recent common ancestor is excluded: <mutations> there would be shared by all sampled <chromosomes> and would not create <segregating sites>.
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