I have to find the following limit:
$$\lim\limits_{n \to \infty} \sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k}$$
I thought I can use something from this other, seemingly similar question, but I don't see any way of manipulating this sum into something easier to work with. So how should I approach this limit?
When $k \in \{0, 1, \dotsc, n\}$ $$ \binom{n}{k} \frac{1}{n2^n+n} \leq \binom{n}{k} \frac{1}{n2^n+k} \leq \binom{n}{k} \frac{1}{n2^n}, $$ whence $$ \frac{2^n}{n(2^n + 1)} \leq \sum_{k = 0}^n \binom{n}{k} \frac{1}{n2^n+k} \leq \frac{2^n}{n2^n}. $$
By squeezing, $$ \lim_{n \to \infty} \sum_{k = 0}^n \binom{n}{k} \frac{1}{n2^n+k} = 0.$$
$$ \lim_{n \to \infty}\sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k} = 0$$
To prove this write $$ \sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k} \le\sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n} = \frac{1}{n} \frac{1}{2^n} \sum\limits_{k = 0}^{n} \binom{n}{k} $$
To see $ \sum\limits_{k = 0}^{n} \binom{n}{k}=2^n$ recall the binomial theorem says $(a+b)^n = \sum_{i=0}^n {n \choose i}a^i b^{n-i}$ for any $a,b \in \mathbb R$. For $a=b=1 $ this becomes $2^n = \sum_{i=0}^n {n \choose i} $. Hence the above gives $$ \sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k} \le \frac{1}{n} \to 0$$
