Computing or approximating $\sum_{n=1}^{N} \log \binom{N}{n} \log (\frac{n+1}{n})$ I'm trying to compute $\sum_{n=1}^{N} \log \binom{N}{n} \log (\frac{n+1}{n})$ when $N$ is large. Is there an asymptotic formula that tells me $\lim_{n \to \infty}  \frac{1}{f(N)} \sum_{n=1}^{N} \log \binom{N}{n} \log (\frac{n+1}{n})$ for some function $f(N)$?

By Stirling's formula
$$
\log \binom{N}{n} \!= n\log \left( {\frac{N}{n}} \right) - (N - n)\log \left( {1 \!-\! \frac{n}{N}} \right) - \frac{1}{2}\log \left( { n\left( {1 \!-\! \frac{n}{N}} \right)} \right) + \mathcal{O}(1),
$$
provided $1\leq n \leq N-1$.
Now
$$
\sum\limits_{n = 1}^{N-1} {\log \left( {n\left( {1 - \frac{n}{N}} \right)} \right)\log \left( {1 + \frac{1}{n}} \right)}  \le \sum\limits_{n = 1}^{N-1} {\log \left( { \frac{N}{4}} \right)\frac{1}{n}}  = \mathcal{O}(\log ^2 N)
$$
and
$$
\sum\limits_{n = 1}^{N-1} {\mathcal{O}(1)\log \left( {1 + \frac{1}{n}} \right)}  \le \mathcal{O}(1)\sum\limits_{n = 1}^{N-1} {\frac{1}{n}}  = \mathcal{O}(\log N).
$$
Also, using $\log \left( {1 + \frac{1}{n}} \right) = \frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)$,
\begin{align*}
\sum\limits_{n = 1}^{N-1} {n\log \left( {\frac{N}{n}} \right)\log \left( {1 + \frac{1}{n}} \right)} & = \sum\limits_{n = 1}^{N-1} {\log \left( {\frac{N}{n}} \right)}  + \mathcal{O}(1)\sum\limits_{n = 1}^{N-1} {\frac{1}{n}\log \left( {\frac{N}{n}} \right)} \\ & = N + \mathcal{O}(\log ^2 N)
\end{align*}
and
\begin{align*}
& \sum\limits_{n = 1}^{N-1} {(N - n)\log \left( {1 - \frac{n}{N}} \right)\log \left( {1 + \frac{1}{n}} \right)} \\ & = N\frac{1}{N}\sum\limits_{n = 1}^{N-1} {\left( {1 - \frac{n}{N}} \right)\frac{N}{n}\log \left( {1 - \frac{n}{N}} \right)}  + \mathcal{O}(1) \sum\limits_{n = 1}^{N-1} {\left( {1 - \frac{n}{N}} \right)\frac{{N }}{{n^2 }}\log \left( {1 - \frac{n}{N}} \right)} \\ & = N\frac{1}{N}\sum\limits_{n = 1}^{N-1} {\left( {1 - \frac{n}{N}} \right)\frac{N}{n}\log \left( {1 - \frac{n}{N}} \right)}  + \mathcal{O}(1)\sum\limits_{n = 1}^{N-1} {\frac{1}{n}}  \\ & = N\int_0^1 {\frac{{(1 - x)\log (1 - x)}}{x}dx}  + \mathcal{O}(1)+\mathcal{O}(\log N) = \left( {1 - \frac{{\pi ^2 }}{6}} \right)N + \mathcal{O}(\log N).
\end{align*}
Thus, in summary,
$$
\sum\limits_{n = 1}^{N-1} {\log \binom{N}{n}\log \left( {1 + \frac{1}{n}} \right)} = \frac{{\pi ^2 }}{6}N + \mathcal{O} (\log^2 N).
$$
With a bit more work, you can improve the error term, for example, into
$$
-\frac{1}{2} \log^2 N +\mathcal{O} (\log N).
$$
