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

Question

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

https://math.stackexchange.com/questions/151441/calculate-sums-of-inverses-of-binomial-coefficients — lab bhattacharjee, Jan 22 '18 at 18:23
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score 5 · Accepted Answer · answered Jan 22 '18 at 18:25

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 $2$.

score 2 · Answer 2 · answered Jan 22 '18 at 18:26

2

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?

answered Jan 22 '18 at 18:26

user296113

7,570

score 1 · Answer 3 · answered Jan 22 '18 at 18:35

$$A_n=\sum_{k=0}^{n}\binom{n}{k}^{-1}=\frac{1}{n!}\sum_{k=0}^{n}\Gamma(k+1)\Gamma(n-k+1)=(n+1)\sum_{k=0}^{n}B(k+1,n-k+1)$$ 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 $2\sum_{k\leq n/2}\binom{n}{k}^{-1}$.

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

3 Answers3