How to calculate this limit

Question

$$\lim_{n\to\infty}\sum_{k = 0}^{n}\left(\frac{k\binom{n}{k} }{n2^n+k}\right) $$

I tried to develop it, but I did not get anything concrete

Answer 1

Hint. Note that $$\sum_{k = 0}^{n}k\binom{n}{k}=D_x\left(\sum_{k = 0}^{n}\binom{n}{k}x^k\right)_{x=1}=D_x\left((1+x)^n\right)_{x=1}=n(1+1)^{n-1}=n2^{n-1}.$$ Therefore $$\frac{n2^{n-1}}{n(2^n+1)}=\frac{1}{n(2^n+1)}\sum_{k = 0}^{n}k\binom{n}{k}\leq \sum_{k = 0}^{n}\frac{k\binom{n}{k} }{n2^n+k}\leq \frac{1}{n2^n}\sum_{k = 0}^{n}k\binom{n}{k}=\frac{1}{2}$$ Now use the Squeeze Theorem.