Sum of Binomials times Logarithms

Question

Is there a closed-form expression or a very good approximation for $$ \sum_{i=0}^n \binom{n}{i} \log (i+1) \,? $$ If the summands alternate, then there is a very close approximation, yet it feels like the alternation is a crucial ingredient.
So I was wondering if one can do better than estimating the sum by $\log(n+1) 2^n$.

Answer 1

The leading asymptotics is not much better than your guess. We have $$\ln(k+1)=\int_0^1\frac{x^k-1}{\ln x}\,dx$$ which gives $$\sum_{k=0}^{n}\binom{n}{k}\ln(k+1)=\int_0^1\frac{(1+x)^n-2^n}{\ln x}\,dx=2^n\big(\ln(n+1)-I_n\big),$$ where \begin{align} I_n&=\int_0^1\left[x^n-\left(\frac{1+x}{2}\right)^n\right]\frac{dx}{\ln x}\\\color{gray}{\left[x=1-\frac{t}{n}\right]\qquad}&=\int_0^n\left[\left(1-\frac{t}{n}\right)^n-\left(1-\frac{t}{2n}\right)^n\right]\frac{dt}{n\ln\left(1-\frac{t}{n}\right)}\\\color{gray}{[\text{apply DCT}]}\qquad&\underset{n\to\infty}{\color{red}{\longrightarrow}}\int_0^\infty\frac{e^{-t/2}-e^{-t}}{t}\,dt=\ln 2. \end{align}

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Detailed asymptotics of $I_n$ can be obtained as follows. Write \begin{align} I_n&=\lim_{\epsilon\to 0}\int\limits_0^{1-2\epsilon}\left[x^n-\left(\frac{1+x}{2}\right)^n\right]\frac{dx}{\ln x}=\lim_{\epsilon\to 0}\left[\int\limits_0^{1-2\epsilon}\frac{x^n\,dx}{\ln x}-\int\limits_{1/2}^{1-\epsilon}\frac{2x^n\,dx}{\ln(2x-1)}\right]\\&=\int\limits_0^{1/2}\frac{x^n\,dx}{\ln x}+\lim_{\epsilon\to 0}\int\limits_{1/2}^{1-\epsilon}x^n\left(\frac{1}{\ln x}-\frac{2}{\ln(2x-1)}\right)dx-\lim_{\epsilon\to 0}\int\limits_{1-2\epsilon}^{1-\epsilon}\frac{x^n\,dx}{\ln x}\\&=\int\limits_0^{1/2}\frac{x^n\,dx}{\ln x}+\int\limits_{1/2}^{1}x^n\left(\frac{1}{\ln x}-\frac{2}{\ln(2x-1)}\right)dx+\ln 2. \end{align} The first integral is $\mathcal{O}(2^{-n})$; we forget it. The second one, after substitution $x=e^{-t}$, is $$-\int_0^{\ln 2}e^{-nt}\varphi(t)\,dt,\qquad\varphi(t)=e^{-t}\left(\frac{1}{t}+\frac{2}{\ln(2e^{-t}-1)}\right).$$ Here, Watson's lemma is applicable; expanding $$\varphi(t)=\frac{1}{2}-\frac{1}{4}t+\frac{1}{6}t^2+\frac{1}{48}t^3+\frac{23}{360}t^4+\frac{31}{480}t^5+\frac{131}{1680}t^6+\ldots,$$ we get $\color{blue}{I_n\asymp\ln 2-\dfrac{1}{2n}+\dfrac{1}{4n^2}-\dfrac{1}{3n^3}-\dfrac{1}{8n^4}-\dfrac{23}{15n^5}-\dfrac{31}{4n^6}-\dfrac{393}{7n^7}}-\ldots$