Limit of sum of terms containing binomial coefficients $$\lim_{n \to \infty} \sum_{k=0}^n \frac{n \choose k}{k2^n+n}$$
The result is $0$. The $n$ from the denominator can be ignored. 
If not for the $k$ at the denominator, the result would be $1$, but I can not find the right inequality.
 A: $$\begin{eqnarray*} \sum_{k=0}^{n}\frac{\binom{n}{k}}{k 2^n+n}\leq \frac{1}{n}+\frac{1}{2^n}\sum_{k=1}^{n}\binom{n}{k}\frac{1}{k}&\stackrel{CS}{\leq}&\frac{1}{n}+\frac{1}{2^n}\sqrt{\zeta(2)\cdot\binom{2n}{n}}\\&\leq&\frac{1}{n}+\frac{1}{2^n}\sqrt{\frac{\zeta(2)\cdot 4^n}{\sqrt{\pi n}}}\\&=&O\left(n^{-1/4}\right).\end{eqnarray*}$$
"CS" stands for the Cauchy-Schwarz inequality and the inequality
$$ \sum_{k=0}^{n}\binom{n}{k}^2 = \binom{2n}{n}\leq\frac{4^n}{\sqrt{\pi n}} $$
is well-known. It comes from the usual manipulations about the Wallis product. 
A: From this P. Stanica paper the approximation
\begin{align}
\binom{n}{k} \sim \sqrt{\frac{2}{n \, \pi}} \, 2^{n} \, e^{-\frac{(n-2k)^{2}}{2n}}
\end{align}
is given. Utilizing this result then the desired limit is as follows:
\begin{align}
\lim_{n \to \infty} \, \sum_{k=0}^{n} \frac{n \choose k}{k2^n+n} &= \lim_{n \to \infty} \, \sqrt{\frac{2}{n \, \pi}} \, \sum_{k=0}^{n} \frac{1}{k+ \frac{n}{2^{n}}} \, e^{-\frac{(n-2k)^{2}}{2n}} \to 0.
\end{align}
