I was playing with a modified version of Pascal’s triangle (with {n \choose k}^{-1} instead of n \choose k everywhere) and this infinite sum popped out:
\sum_{k=2}^{\infty}\sum_ {n=1}^{\infty} \frac{1}{n(n+1)(n+2)…(n+k-1)}
The partial sums seem to approach \alpha \approx 1.317…
Does a closed form for \alpha exist?
Answer
It is not difficult to de-nest such double series.
Lemma 1. For any k\geq 2, we have
\sum_{n\geq 1}\frac{1}{n(n+1)\cdot\ldots\cdot(n+k-1)}=\sum_{n\geq 1}\frac{1}{(n)_k}=\frac{1}{(k-1)\cdot(k-1)!}.
Proof: it is enough to exploit Euler’s beta function and a geometric series, since:
\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{(n)_k}=\sum_{n\geq 1}\frac{\Gamma(n)}{\Gamma(n+k)}&=&\frac{1}{\Gamma(k)}\sum_{n\geq 1}B(k,n)\\&=&\frac{1}{\Gamma(k)}\sum_{n\geq 1}\int_{0}^{1}x^{n-1}(1-x)^{k-1}\,dx\\&=&\frac{1}{\Gamma(k)}\int_{0}^{1}\sum_{n\geq 1}(1-x)^{k-1}x^{n-1}\,dx\\&=&\frac{1}{\Gamma(k)}\int_{0}^{1}(1-x)^{k-2}\,dx\\&=&\frac{1}{\Gamma(k)}\int_{0}^{1}x^{k-2}\,dx = \frac{1}{(k-1)\cdot(k-1)!}.\end{eqnarray*}
In particular, we have:
S = \sum_{k\geq 2}\sum_{n\geq 1}\frac{1}{(n)_k} = \sum_{m\geq 1}\frac{1}{m\cdot m!}=\int_{0}^{1}\sum_{m\geq 1}\frac{x^{m-1}}{m!}\,dx = \color{red}{\int_{0}^{1}\frac{e^x-1}{x}\,dx}.
The latter is not an elementary integral but an exponential integral.
At last, a huge WELCOME to MSE.
Attribution
Source : Link , Question Author : Francesco Baccetti , Answer Author : Jack D’Aurizio
