How to estimate (approximate) $\sum _{k=1}^{n}{\log(k)\binom {n}{k}}$?

Question

I am looking for an estimation or an approximation of

$\sum _{k=1}^{n}{\log(k)\binom {n}{k}}$

Any hints will be appreciated. Thank you.

Answer 1

For $\log n \ge 2$ $$\sum_{k=1}^n {n \choose k} \log(k) \ge \sum_{k=n/\log n}^n {n \choose k} \log(n/\log n)$$ $$=\sum_{k=1}^n {n \choose k} \log(n/\log n)-\sum_{k=1}^{n/\log n} {n \choose k} \log(n/\log n)$$ $$ \ge \sum_{k=1}^n {n \choose k} \log(n/\log n)-\frac{n/\log n}{n}\sum_{k=1}^n {n \choose k} \log(n/\log n)$$ $$ = \log(n/\log n)2^n- \log(n/\log n) 2^n/\log n$$ $$ = \log(n) 2^n(1-\frac1{\log n})(1-\frac{\log \log n}{\log n})$$ Together with the obvious bound $$\sum_{k=1}^n {n \choose k }\log(k) \le \sum_{k=1}^n {n \choose k }\log(n)=\log(n) 2^n$$ we get $$\frac{\sum_{k=1}^n {n \choose k }\log(k)}{\log(n) 2^n} \in [(1-\frac1{\log n})(1-\frac{\log \log n}{\log n}),1]$$

Answer 2

I'm continuing the computations by Jack D'Aurizio (following myself). \begin{align}\sum_{k=1}^{n}\binom{n}{k}\ln k&=\sum_{k=1}^{n}\binom{n}{k}\int_0^1\frac{x^{k-1}-1}{\ln x}\,dx\\\color{gray}{[\text{note the sign}]}\quad&=\int_0^1\frac{(1+x)^n-1-(2^n-1)x}{x\ln x}\,dx\\\color{gray}{\text{[integrate by parts]}}\quad&=\int_0^1\big(2^n-1-n(1+x)^{n-1})\ln\ln\frac{1}{x}\,dx\\\color{gray}{[\text{substitute }x=2e^{-t}-1]}\quad&=-(2^n-1)\gamma-2^n n\int_0^{\ln 2}e^{-nt}\ln\ln\frac{1}{2e^{-t}-1}\,dt\\&=\color{blue}{2^n(\ln n-\ln 2-I_n)}+\underbrace{\gamma+2^n n\int_{\ln 2}^{\infty}e^{-nt}\ln 2t\,dt}_{\text{small, can be neglected}},\end{align} $$I_n=n\int_0^{\ln 2}e^{-nt}\varphi(t)\,dt,\qquad\varphi(t)=\ln\left[\frac{1}{2t}\ln\frac{1}{2e^{-t}-1}\right].$$ $I_n$ fits Watson's lemma: $\varphi(t)=\frac{1}{2}t+\frac{3}{8}t^2+\frac{1}{3}t^3+\frac{65}{192}t^4+\frac{67}{180}t^5+\ldots$ gives $$I_n\asymp\frac{1}{2n}+\frac{3}{4n^2}+\frac{2}{n^3}+\frac{65}{8n^4}+\frac{134}{3n^5}+\ldots$$

Answer 3

Interesting question. Frullani's integral provides an integral representation for $\log(k)$, $$ \log(k) = \int_{0}^{+\infty}\frac{e^{-x}-e^{-kx}}{x}\,dx $$ which in combination with the binomial theorem leads to $$ \begin{eqnarray*}\sum_{k=1}^{n}\log(k)\binom{n}{k} &=& \int_{0}^{+\infty}\left[(2^n-1)e^{-x}-(1+e^{-x})^n+1\right]\frac{dx}{x}\\&=&\int_{0}^{1}\underbrace{\frac{1-(1+t)^n+t(2^n-1)}{t\log(t)}}_{f_n(t)}\,dt.\end{eqnarray*} $$ Most of the mass of the integral comes from the right endpoint of the integration range, and since $$ \lim_{t\to 1^-}f_n(t) = 1+\left(\frac{n}{2}-1\right)2^n,\qquad \lim_{t\to 1^-}f_n'(t) = -\frac{1}{2}+\left(\frac{n^2}{8}-\frac{3n}{8}+\frac{1}{2}\right)2^n $$ we may approximate $$ \int_{0}^{1}f_n(t)\,dt \approx \int_{0}^{1} f_n(1^-)e^{-\frac{f_n'(1^-)}{f_n(1^-)}x}\,dx\approx \frac{2^{n+1}(n-2)^2}{n^2-3n+4}\left(1-e^{-\frac{n^2-3n+4}{4n-8}}\right). $$