Asymptotic approximation of the expression $\sum _{i=0}^n \log (\binom{n}{i}+1)$

I am trying to work out the asymptotic approximation (for large $n$) of the following: $$\sum _{i=0}^n \log (\binom{n}{i}+1)$$ So far I tried to write the expression as, $$\sum^{n}_{i=0}log(x)=-\sum^{n}_{i=0}x\int^{\infty}_{0}e^{-xt}\log tdt-\gamma$$ where I took $x=\binom{n}{i}+1$ and $\gamma$ is Euler gamma, but I am not sure if this method is so useful. Any other approach is appreciated.

• Is the $+1$ (under the logarithm) essential for the result you need? What is the precision you require? – metamorphy Aug 28 '18 at 14:51
• (For the main term you can just drop it and apply the Stirling's formula.) – metamorphy Aug 28 '18 at 14:55
• @metamorphy so if one drops 1, then one can write the sum in terms of Gamma and Barnes G functions. I was wondering about keeping +1 to have as much as precision as possible. – William Aug 28 '18 at 15:24

$$S:=\sum_{k=0}^{n}\log(1+\binom{n}{k}) =\sum_{k=0}^{n} \log(\binom{n}{k}(1+1/{\binom{n}{k}})\sim$$ $$\sim\sum_{k=0}^{n} \log(\binom{n}{k})+\sum_{k=0}^n\frac{1}{\binom{n}{k}}-\frac{1}{2}\sum_{k=0}^{n}\frac{1}{\binom{n}{k}^2}+...$$ where a Taylor series of $\log(1+x)$ has been used. It can be shown that $$\sum_{k=0}^n\frac{1}{\binom{n}{k}} =2(1+\frac{1}{n}+...) \text{ and } \sum_{k=0}^n\frac{1}{\binom{n}{k}^2}=2(1+1/n^2+...)$$ We will always know the leading coefficient is 2 because of the $k=0$ and $k=n$ terms. We have an infinite sum of these terms, which fortunately converges because of their alternating weight of $(-1)^m/m.$ The log-binomial sum can be done in closed form: $$\sum_{k=0}^{n} \log(\binom{n}{k})=\log\Big(\frac{n!^{n+1}}{G(n+2)^2}\Big).$$ $G$ is the Barne's G and its asymptotic expansion is well-known. Collecting, $$S\sim \log\Big(\frac{n!^{n+1}}{G(n+2)^2}\Big) + 2\log{2}+\frac{2}{n}+O(\frac{1}{n^2})$$ Use this expansion for $n=20$ and 4 correct digits are obtained.