Show that $\sum^{N-1}_{i=1} \frac{N-1}{i(N-i)}$ is approximately $2\log N$ for large $N$.

Question

I am working on a problem relating to a Markov chain, that results in the sum:

$$\sum^{N-1}_{i=1} \frac{N-1}{i(N-i)}$$

For the expected time until reaching the $N^{th}$ state from $0$.

The question asks to show that for large N, the expected time is approximately $2\log{N}$, but I'm not sure how to get this from the above sum. Is there an identity that leads to that result?

Answer 1

Turn it into an integral $$\sum^{N-1}_{i=1} \frac{N-1}{i(N-i)} \approx \int_1^{N}\frac {N-1}{x(N-x)}dx$$ Now do partial fractions on the integral $$\int_1^{N}\frac {N-1}{x(N-x)}dx=\int_1^N\left(\frac {N-1}{Nx}+\frac {N-1}{N(N-x)}\right)dx$$ and each term integrates to $\frac {N-1}N\log N$, the second by recognizing it is the same integral as the first done backwards.