Funny double infinite sum
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?
Solution 1:
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.
Solution 2:
Swapping the order of summation we see the reciprocals of OEIS sequence A001563, i.e. $\sum_{n\ge 1} \frac{1}{n\cdot n!}$. Alpha tells us the sum is $$Ei(1)-\gamma$$
Ei is the exponential integral, and $\gamma$ is the Euler-Mascheroni constant. Of course, this is sort of circular, since $Ei(1)$ is basically the series we're trying to compute, up to a constant $\gamma$ difference.
This is probably as "closed form" as you're going to find.
Solution 3:
[Not an answer, but too long to fit as a comment.]
[Edit: @adjan points out that this is known as the Leibniz harmonic triangle, which I was unaware of.]
I don't know if this is related or not: I noticed a curious fact a few years ago about reciprocals of binomial coefficients. If you take Pascal's triangle, but instead of putting $\binom{n}{k}$ in each entry, you put the reciprocal of $\,(n+1)\binom{n}{k}, $ you get an upside-down Pascal's triangle, with each number being the sum of the two numbers below it:
\begin{array} \\&&&&&1 \\&&&&\frac12&&\frac12 \\&&&\frac13&&\frac16&&\frac13 \\&&\frac14&&\frac1{12}&&\frac1{12}&& \frac14 \\& \frac15&&\frac1{20}&&\frac1{30}&&\frac1{20}&&\frac15 \\\ .^{\large{.}^{\LARGE{.}}}&&\vdots&&\vdots&&\vdots&&\vdots&&{}^{{}^{{}^{\LARGE{.}}}}{}^{\hspace{-1mu}\large{.}}. \end{array} $$ $$ The proof that it works is straightforward: \begin{align}\require{cancel} \frac1{(n+1)\binom{n}{k}}+\frac1{(n+1)\binom{n}{k+1}}&=\frac{1}{n+1}\frac{\binom{n}{k}+\binom{n}{k+1}}{\binom{n}{k}\binom{n}{k+1}} \\&=\frac1{n+1}\binom{n+1}{k+1}\frac{k!\,(n-k)!}{n!}\frac{(k+1)!\,(n-k-1)!}{n!} \\&=\frac1{n+1}\frac{(n+1)!}{\bcancel{(k+1)!}\cancel{(n-k)!}}\frac{k!\,\cancel{(n-k)!}}{n!}\frac{\bcancel{(k+1)!}\,(n-k-1)!}{n!} \\&=\frac{\cancel{(n+1)!}}{\cancel{(n+1)\cdot n!}}\frac{k!\,(n-k-1)!}{n!} \\&=\frac1{n\binom{n-1}{k}}. \end{align}
I have no idea if this is well-known or not — I hadn't come across it before.