Is it true that $\sum_{k=0}^{n}k\cdot \left(\begin{array}{l}{n}\\{k}\end{array}\right)=O\left(2 ^ {n\log _{3}n}\right)?$

Question

Problem: Is it true that $\sum_{k=0}^{n}k\cdot \left(\begin{array}{l}{n}\\{k}\end{array}\right)=O\left( 2 ^ {n\log _{3}n}\right)?$

My start of solution: $$\sum_{k=0}^{n}k\cdot \left(\begin{array}{l}{n}\\{k}\end{array}\right)\leq \sum_{k=0}^{n}k\cdot \left(\begin{array}{l}{n}\\{\lfloor \frac{n}{2}\rfloor}\end{array}\right)\leq \frac{n\cdot(n+1)}{2}\cdot \left(\begin{array}{l}{n}\\{\lfloor \frac{n}{2}\rfloor}\end{array}\right)\leq n(n+1)! \leq nn^n \leq n^{n+1}$$

I think this upper bound is way too large and I can't seem to find a solution.

Answer 1

$$\sum_{k=0}^n k\binom nk=n2^{n-1}=2^{n-1+\log_2n}.$$ How do $n-1+\log_2n$ and $n\log_3 n$ compare?