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

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 
$$\frac{\sum_{k=1}^n {n \choose k }\log(k)}{\log(n) 2^n} \in [(1-\frac1{\log n})(1-\frac{\log \log n}{\log n}),1]$$
