Is this "combinatorial" sum equal to $1$ for every natural $m$?

Question

I did some computations and it seems to me that this holds:

$\dfrac {m}{m+1} \cdot \sum_{k=1}^{\infty} \dfrac {1}{m+k \choose m+1}=1$ for every $m \in \mathbb N$.

How to prove this beautiful identity, if it is really true?

Answer 1

It is enough to exploit Euler's Beta function and a geometric series: $$\begin{eqnarray*}\sum_{k\geq 1}\binom{m+k}{m+1}^{-1} &=& \sum_{k\geq 1}\frac{(m+1)!(k-1)!}{(m+k)!}\\ &=& (m+1)\sum_{k\geq 1}\frac{\Gamma(m+1)\Gamma(k)}{\Gamma(m+k+1)}\\ &=& (m+1) \sum_{k\geq 1} B(k,m+1) \\ &=& (m+1) \sum_{k\geq 1}\int_{0}^{1} x^{k-1}(1-x)^{m}\,dx\\ &=& (m+1) \int_{0}^{1}(1-x)^{m-1}\,dx\\ &=& (m+1) \int_{0}^{1} x^{m-1}\,dx = \color{blue}{\frac{m+1}{m}}. \end{eqnarray*}$$

Answer 2

The proof that this identity holds is easy if you are familiar with Discrete Calculus and negative-exponent falling powers.

In particular, $$\frac{1}{\binom{m+k}{m+1}} = \frac{(m+1)!}{(m+k)^{\underline{m+1}}} = (m+1)! k^{\underline{-(m+1)}} $$

For example, with $m=3,k=7$, $$ \frac{1}{\binom{10}{4}} = \frac{4!}{7\cdot 8\cdot 9\cdot 10} $$

The rules of finite summation work much like those of integration. The "indefinite sum" is $$ \sum x^{\underline{n}} = \frac{1}{n+1} x^{\underline{n+1}} $$ (except if $n=-1$ in which case the analogue of $\int x^{-1} dx = \log x$ is $\sum n^{-1} = H_n$, the harmonic sum function).

In particular $$ \sum k^{\underline{-(m+1)}} = -\frac{1}{m} k^{\underline{-m}} $$ and inserting the limits $1$ to $\infty$ the infinity end gives zero and the minus sign goes away leaving $$ \sum_{k=1}^\infty k^{\underline{-(m+1)}} = \frac{1}{m} \cdot 1^{\underline{-m}} = \frac{1}{m\cdot m!}$$

So $$\frac{1}{\binom{m+k}{m+1}} =(m+1)!\sum \frac{1}{(m+k)^{\underline{m+1}}} = (m+1)! \frac{1}{m\cdot m!} = \frac{(m+1)!}{m!} \frac{1}{m} = \frac{m+1}{m}$$

Answer 3

I hope noone minds, if I add an easy approach to this after such a long time. I really enjoyed learning something here from previous answers, but thought a basic approach might be helpful or interesting as well. $$\sum_{k=1}^n \frac{1}{\binom{m+k}{m+1}}$$$$=\sum_{k=1}^n\frac{(k-1)!(m+1)!}{(m+k)!}$$$$=(m+1)!\sum_{k=1}^n\frac{1}{k\cdot\ldots\cdot(m+k)}$$$$=\frac{(m+1)!}{m}\sum_{k=1}^n\frac{m+k-k}{k\cdot\ldots\cdot(m+k)}$$$$=\frac{(m+1)!}{m}\sum_{k=1}^n\frac{1}{k\cdot\ldots\cdot(m+k-1)}-\frac{1}{(k+1)\cdot\ldots\cdot(m+k)}$$$$=\frac{(m+1)!}{m}(\frac{1}{m!}- \frac{1}{(n+1)\cdot\ldots\cdot(n+k)})$$ and letting n go to infinity brings the result.