Joy of Mathematics

In this post we are discussing the sum

$S_n^{(m)} =\sum_{k=0}^{n} k^m \binom{n}{k}^{-1}$

In $1947$ Staver was the first to study the sum $S_n^{(0)}$ . He observed that $\displaystyle S_n^{(1)} = \frac{n}{2} S_n^{(0)}$ . He was , then , able to extract the recursive relation

(1) $\begin{equation*} S_{n+1}^{(0)} = \frac{n+2}{2(n+1)} S_n^{(0)} +1 \end{equation*}$

He then proved a great result which is well known in literature

(2) $\begin{equation*} {\colorbox{cyan}{\color{DarkRed}\boxed{\rm{\sum_{j=0}^{n}\binom{n}{j}^{-1}=\frac{n+1}{2^{n+1}}\sum_{j=1}^{n+1}\frac{2^j}{j}}}}} \end{equation*}$

Later, in $1981$ Rocket combining the identity

(3) $\begin{equation*} \binom{n}{k}^{-1} = \binom{n-1}{k-1}^{-1} - \frac{n-k}{n-k+1} \binom{n}{k-1}^{-1} \end{equation*}$

along with induction he was able to give another proof of $(2)$ . In $1993$ Surin provided another proof using the well known integral representation of the binomial coefficient,

(4) $\begin{equation*} \binom{n}{k}^{-1} = (n+1)\int_{0}^{1} x^k \left ( 1-x \right )^{n-k} \, {\rm d}x \end{equation*}$

Since then the cases $m=0$ and $m=1$ have been studied extensively. However, Mansour generalising the idea of Sury provided a theorem which states the following:

Theorem [Mansour]: Let $r,n\geq k$ be non negative integers and $f(n, k)$ be given by

$f(n, k) = \frac{(n+r)!}{n!} \int_{u_1}^{u_2} p^k(t) q^{n-k}(t) \, {\rm d}t$

where $p, q$ are two functions defined on $[u_1, u_2]$ . Let $\{a_n\}_{n \in \mathbb{N}}$ , $\{b_n\}_{n \in \mathbb{N}}$ be two sequences and $A(x)$ , $B(x)$ be their corresponding generating functions. Then,

$\displaystyle \sum_{n=0}^{\infty} \left ( \sum_{k=0}^{n} f(n, k) a_k b_{n-k} \right ) x^n = \frac{\mathrm{d}^r}{\mathrm{d} x^r} \left ( x^r \int_{u_1}^{u_2} A\left ( xp(t) \right ) B(xq(t))\, {\rm d}t \right )$

Proof: The proof is a standard generating type one and is left to the reader.

Using the above theorem along with equation $(4)$ we can generate wonderful stuff. For example:

Example 1: Pick $a_n=a^n$ and $b_n=b^n$ . Then

$\begin{aligned} \sum_{n=0}^{\infty} \left ( \sum_{k=0}^{n} a^k b^{n-k} \binom{n}{k}^{-1} \right )x^n &= \frac{\mathrm{d} }{\mathrm{d} x} \left ( x \int_{0}^{1} \frac{{\rm d}t}{\left ( 1-axt \right )\left ( 1-bx+bxt \right )} \right ) \\ &= \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{-\log (1-ax)-\log(1-bx)}{a+b-abx} \right ) \end{aligned}$

which , after a bit of transformations gives the general result

${\colorbox{cyan}{\color{DarkRed}\boxed{\sum_{k=0}^{n}a^n b^{n-k} \binom{n}{k}^{-1} =\frac{n+1}{\left ( a+b \right )\left ( \frac{1}{a}+\frac{1}{b} \right )^{n+1}} \sum_{k=1}^{n+1} \frac{\left ( a^k+b^k \right )\left ( \frac{1}{a}+ \frac{1}{b} \right )^{k}}{k}}}}}$

Fabulous, isn’t it? If we set $a=b=1$ we get equation $(2)$ . Of course there are other applications of the above theorem. We can establish a similar equality for the sum $S_n^{(1)}$ . Another relation that can be established by the above theorem is the following:

(5) $\begin{equation*}{\colorbox{cyan}{\color{DarkRed} \boxed{\sum_{k=0}^{n} \binom{2n}{2k}^{-1} = \frac{n(2n+1)}{2^{2n+2}} \sum_{k=0}^{2n+1} \frac{2^k}{k+1}}}} \end{equation*}$

We are not gonna go into a deep analysis but the following equalities also hold:

(6) $\begin{equation*} \binom{n}{k}^{-m} =\left [\left ( n+1 \right )^m \int_{0}^{1}t^k (1-t)^{n-k}\, {\rm d}t \right ]^m \end{equation*}$

and

(7) $\begin{equation*} \sum_{k=0}^{n} \binom{n}{k}^{-m} =(n+1)^m \sum_{k=0}^{n}\left [ \sum_{i=0}^{k} \frac{(-1)^i}{n-k+1+i} \binom{k}{i} \right ]^m \end{equation*}$

Tag: Calculus

On the sum of inverse binomial