TITLE: Funny double infinite sum
QUESTION [57 upvotes]: 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?

REPLY [15 votes]: [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.