Sum Fun

First we'll look at $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{\dbinom{k + n}{k}} = \sum_{k=1}^{\infty} \dfrac{1}{\dbinom{k + n}{n}} =$

$\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{\dfrac{(k + 1)(k + 2) \cdots (k + n)}{n!}} =$

$n! \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{(k + 1)(k + 2) \cdots (k + n)} =$

$\dfrac{n!}{n - 1} \displaystyle\sum_{k=1}^{\infty} \left( \dfrac{1}{(k + 1)(k + 2) \cdots (k + n - 1)} - \dfrac{1}{(k + 2)(k + 3) \cdots (k + n)} \right)=$

$\dfrac{n!}{n - 1} ( \dfrac{1}{2 * 3 \cdots n} - \dfrac{1}{3 * 4 \cdots (n + 1)} + \dfrac{1}{3 * 4 \cdots (n + 1)} -$

$\dfrac{1}{4 * 5 \cdots (n + 2)} + \dfrac{1}{4 * 5 \cdots (n + 2)} -+ \cdots ) =$

$\dfrac{n!}{n - 1} * \dfrac{1}{n!} = \dfrac{1}{n - 1}$ ,

where in the next to last step we have formed a telescoping series with all the terms in the brackets canceling out in pairs except the first term. So now

$S = \displaystyle\sum_{n=2}^{\infty} \dfrac{(-1)^{n}}{n - 1} = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} +- \cdots = \ln(2) = 0.693147181.....$

Thus $\lfloor 10000*S \rfloor = \boxed{6931}.$