In the same way that $\sum\limits_{n=1}^\infty n = -\frac{1}{12}$ (Yes, I know it doesn't really! But lets say the series can be assigned the value $-1/12$ for example by continuing the zeta function to -1.)
In a similar way, I want to work out the value assigned to:
$f(d) = \sum\limits_{n=0}^\infty \frac{(d+n)!}{n!d!} = (1 + \frac{(d+1)}{1!} + \frac{(d+1)(d+2)}{2!} + ...)$
when $d$ is an integer.
Is there a way to assign a value to this function? (Note that when $d=1$ the answer should be $f(1)=-1/12$ since then it becomes $1+2+3+...$.)
For example $f(2) = 1 + 3 + 6 + 10 +...$ an infinite sum of triangle numbers.
Is it possible?
With $m = n+1$, we write $$ \dfrac{(d+m-1)!}{(m-1)!\; d!} = \sum_{j=1}^d \dfrac{(-1)^{d-j} S_{d,j} m^j}{d!} $$ where $S_{d,j}$ is a Stirling number of the first kind. Thus for a Zeta function regularization we could write (formally) $$ \sum_{m=1}^\infty \dfrac{(d+m-1)!}{(m-1)!\; d!} m^{-s} = \sum_{m=1}^\infty \sum_{j=1}^d \dfrac{(-1)^{d-j} S_{d,j} m^j}{d!} m^{-s} = \sum_{j=1}^d \dfrac{(-1)^{d-j} S_{d,j}}{d!} \zeta(s-j)$$ and take $s=0$, obtaining $$ f(d) = \sum_{j=1}^d \dfrac{(-1)^{d-j} S_{d,j}}{d!} \zeta(-j)$$ The first few values, for $d$ from $1$ to $10$, are $$ -\frac{1}{12},-\frac{1}{24},-{\frac{19}{720}},-{\frac{3}{160}},-{\frac{863}{60480}},- {\frac{275}{24192}},-{\frac{33953}{3628800}},-{\frac{8183}{1036800}},- {\frac{3250433}{479001600}},-{\frac{4671}{788480}} $$
Hmm: it looks like:
$$ f(d) = \dfrac{1}{(d+1)!} \int_0^1 dx \; \prod_{j=0}^d (j-x) $$
which is related to Bernoulli numbers of the second kind.