Asymptotic behaviour of mean absorption time at the Moran model

Question

In the Moran model, a model from population genetics, the mean time until absorption is given by

$$ \tau _i =N \left( \sum_{j=1}^i \frac{N-i}{N-j} + \sum_{j=i+1}^{N-i} \frac{i}{j} \right),$$

where $N$ is a natural number greater than $2$ and $i\in \{1,\ldots,N-1\}$. We are interested in the asymptotic behaviour of $\tau_i$ if $N \rightarrow \infty$ and $i/N \rightarrow x \in (0,1)$. In particular, we want to show that

$$ \frac{\tau_i}{N^2} \rightarrow -(1-x) \log(1-x) -x \log(x) := \tau(x). $$

As a hint we should use the fact that $$ \lim_{n \rightarrow \infty} \ \sum_{k=1}^{n} 1/k - \log(n) = \gamma, $$

where $\gamma$ is the Euler-Mascheroni constant. So far I could not show the above statement. Maybe I somehow need to use some Taylor expansion of $\tau(x)$ but unfortunately my ideas are very vague. Thanks for any advice!

Answer 1

$$\tau _i =N \left( \sum_{j=1}^i \frac{N-i}{N-j} + \sum_{j=i+1}^{N-i} \frac{i}{j} \right)\\ =N^2\left((1-x)\sum_{j=1}^{i}\frac1{N-j}+x\sum_{j=i+1}^{N-i}\frac1j\right)\\ =N^2\left((1-x)\sum_{j=N-Nx}^{N-1}\frac1j+x\sum_{j=i+1}^{N-Nx}\frac1j\right)\\ \approx N^2\left((1-x)\left[\log(N)+\gamma-\log(N-Nx)-\gamma\right]+x\left[\log(N-Nx)+\gamma-\log(Nx)-\gamma\right]\right)\\ \approx N^2\left(-(1-x)\log(1-x)+x\log((1-x)/x)\right)$$ Should the $N-i$ in the second sum be $N-1$?

Answer 2

Do these formulas help?

$$ \log\left(\frac{p}{q}\right)=\sum_{i=0}^\infty \left(\sum_{j=pi+1}^{p(i+1)}\frac{1}{j}-\sum_{k=qi+1}^{q(i+1)}\frac{1}{k}\right) $$ http://oeis.org/A166871

$$ \gamma= \sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right) $$

Euler-Mascheroni constant expression, further simplification