Finding sum of $\lim_{n\rightarrow \infty}\sum^{n}_{k=0}\binom{n}{k}^{-1}$ 
Finding sum of $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\binom{n}{k}^{-1}$

Try: $$\lim_{n\rightarrow\infty}\sum^{n}_{k=0}\frac{k!\cdot (n-k)!}{n!}$$
$$\lim_{n\rightarrow}(n+1)\sum^{n}_{k=0}\int^{1}_{0}x^{n-k}(1-x)^kdx$$
$$\lim_{n\rightarrow\infty}\int^{1}_{0}x^{n}\sum^{n}_{k=0}\left(\frac{1-x}{x}\right)^{k}dx$$
$$\lim_{n\rightarrow\infty}\int^{1}_{k=0}x^{k}\sum^{n}_{k=0}\frac{x^{k+1}-(1-x)^{k+1}}{(2x-1)(1-x)^k}dx$$
Could some help me to solve it, Thanks
 A: The sum contains:


*

*Two terms that are equal to $\dbinom{n}{0}^{-1} = \dbinom{n}{n}^{-1} = 1$

*Two terms that are equal to $\dbinom{n}{1}^{-1} = \dbinom{n}{n-1}^{-1} = \dfrac{1}{n}$

*$n-3$ terms that are $\le \dbinom{n}{2}^{-1} = \dbinom{n}{n-2}^{-1} = \dfrac{2}{n(n-1)} = O(n^{-2})$


This is enough to show that for large $n$, the sum is $2+O(n^{-1})$. Therefore, the limit is $2$.
A: We have
$$\sum^{n}_{k=0}\left(\binom{n}{k}\right)^{-1}=2+\frac2n +\sum^{n-2}_{k=2}\left(\binom{n}{k}\right)^{-1}$$
where in the last sum there are $n-3$ terms less than $\frac C{n^2}$ for some constant $C$. Can you take it from here?
A: $$A_n=\sum_{k=0}^{n}\binom{n}{k}^{-1}=\frac{1}{n!}\sum_{k=0}^{n}\Gamma(k+1)\Gamma(n-k+1)=(n+1)\sum_{k=0}^{n}B(k+1,n-k+1)$$
hence
$$ A_n = (n+1)\int_{0}^{1}\sum_{k=0}^{n} x^{n-k} (1-x)^{k}\,dx\stackrel{\text{symmetry}}{=}2(n+1)\int_{0}^{1/2}\frac{(1-x)^{n+1}-x^{n+1}}{1-2x}\,dx $$
where $g_n(x)=\frac{(1-x)^{n+1}-x^{n+1}}{1-2x}$ is a positive and decreasing function which smoothly goes from $1$ to $(n+1)2^{-n}\approx 0$ over the interval $\left(0,\frac{1}{2}\right)$. We may apply Laplace's method by noticing that $g_n(x)\approx e^{-(n-1)x}$, such that
$$ \lim_{n\to +\infty} A_n = \lim_{n\to +\infty}2(n+1)\int_{0}^{1/2}e^{-(n-1)x}\,dx =\color{red}{2}$$
as conjectured. Another approach is just to invoke the dominated convergence theorem applied to $2\sum_{k\leq n/2}\binom{n}{k}^{-1}$.
