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 $f(N)$?
 A: 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
$$
-\frac{1}{2} \log^2 N +\mathcal{O} (\log N).
$$
A: Notice that for $n \geq 1$
$$\frac{\log(2)}{n}\leq\log(1+\frac1{n})\leq\frac{\log(3)}{n}$$
So all you need is estimation for
$$\sum_{n=1}^{N}\frac{\log{N \choose n}}{n}$$
You can use next
$$ \frac{N^n}{n^n} \leq {N \choose n} < \frac{(eN)^n}{n^n}$$
giving
$$\sum_{n=1}^{N}\frac{N^n}{n^{n+1}} \leq \sum_{n=1}^{N}\frac{\log{N \choose n}}{n}<\sum_{n=1}^{N}\frac{(eN)^n}{n^{n+1}}$$
So all you need is an estimation for
$$\sum_{n=1}^{N}\frac{(kN)^n}{n^{n+1}}, 1 \leq k < e $$
Easily switch to an integral first
$$ \int \frac{a^x}{x^{x+1}} dx$$
In order to get the estimation start from the opposite
$$ F(x) = \int \frac{a^x}{x^{x+1}} dx$$
$$ F'(x) = \frac{a^x}{x^{x+1}}$$
Take $F(x)=\frac{a^x}{x^{x+1}}g(x)$
Now differentiate and compare
$$\frac{a^x}{x^{x+1}}=\frac{a^x}{x^{x+1}}(g(x)(\log(a)-1-\log(x)-\frac1{x})+g'(x)$$
meaning it is the best to take
$$g(x)=\frac1{\log(a)-1-\log(x)-\frac1{x}}$$
which is giving
$$F(x)=\frac{a^x}{x^{x+1}(\log(a)-1-\log(x)-\frac1{x})}$$
From this you get
$$ \int_{1}^{N} \frac{(kN)^x}{x^{x+1}} dx \sim \frac{k^N}{N(\log(k)-1)-1}$$
Obviously $k$ cannot be $1$.
You can use the same idea with various other estimation for ${N \choose n}$. This is just an illustration.
A: Not a complete answer, but a helpful one to understand the asymptotic behavior. Write $g(N) = \sum_{n=1}^{N} \log \binom{N}{n} \log(\frac{n+1}{n}).$ Write $h(N) = g(N+1)-g(N)$ and note that $$h(N) =  \log \binom{N+1}{n+1} \log(\frac{N+2}{N+1})  + \sum_{n=1}^{N} (\log  \binom{N+1}{n} - \log \binom{N}{n}) \log \frac{n+1}{n} =$$
$$=  0 + \sum_{n=1}^{N} (\log \frac{\binom{N+1}{n}}{ \binom{N}{n}} ) \log \frac{n+1}{n} $$
$$ = \sum_{n=1}^{N} \log(\frac{N+1}{N+1-n})  \log \frac{n+1}{n} .$$
If $g(N)$ grew asymptotically linearly, then $h(N) \to \Omega(1)$. If $g(N)$ grew quadratically, then $h(N) \to \Omega(N)$.
I ran this through Mathematica and got
$$h(N) = \frac{\log ^2 \left(N^2 H_N+N H_N+H_N+N^3+N^2+\gamma  N+N \psi ^{(0)}(N+1)+2 \psi
   ^{(0)}(N+1)+2 \gamma \right)}{N+1}$$
where $\psi^{(0)}$ is the Digamma function. This seems to go to 0 as $N$ grows to infinity although Mathematica tells me it grows at a rate $\log^2(N)$. Either way, it seems $g(N)$ grows much slower than quadratic.
