Closed form for $\sum_{k=0}^{n} k\binom{n}{k}\log\binom{n}{k}$

Is it possible to write this in closed form: $$\sum_{k=0}^{n} k\binom{n}{k}\log\left(\vphantom{\Huge A}\binom{n}{k}\right)$$

Can you get something like $$n2^{n-1}\log(2^{n-1})$$

Solution 1:

Warning!

I couldn't find a closed form. An approximation is described below.

You may start by symmetrizing the summand to get $$\sum_{k=0}^{n} k\binom{n}{k}\log\binom{n}{k}={n\over 2}\sum_{k=0}^{n} \binom{n}{k}\log\binom{n}{k}.\tag1$$

The terms in the sum on the right hand side of (1) are symmetric around $n/2$ and concentrated near $k\approx n/2$, so replacing $\log{n\choose k}$ with $\log{n\choose n/2}$ gives a reasonable approximation, and an upper bound. That is, $${n\over 2}\sum_{k=0}^{n} \binom{n}{k}\log\binom{n}{k}\approx {n\over 2}\,2^n\log{n\choose n/2}.$$

Using Stirling's formula gives another approximation (and upper bound) $${n\over 2} \sum_{k=0}^{n} \binom{n}{k}\log\binom{n}{k}\approx {n\over 2}\,2^n [(n+1/2)\log(2)-\log(n\pi)/2].$$

Added: A better approximation results by replacing $\log{n\choose k}$ with $\log{n\choose n/2}-{2\over n}(k-n/2)^2$. With a little work you can get $${n\over 2}\,\sum_{k=0}^{n} \binom{n}{k}\log\binom{n}{k}={n\over 2}\,2^n \left[\log{n\choose n/2}-{1\over 2}+o(1)\right].$$