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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?

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    $\begingroup$ Oh dear not the 1st sum again. $\endgroup$ – StubbornAtom Sep 7 '16 at 18:34
  • $\begingroup$ $\sum\limits_{n=1}^\infty n = -\frac{1}{12}$ has striken again ! The least you can say about this mock result is cite at least one source... And if I have an advice do give you : don't waste your youth in such deadends... $\endgroup$ – Jean Marie Sep 7 '16 at 18:38
  • $\begingroup$ I understand what you're saying but your question is essentially: 'can we assign a wrong value to $f(d)$ in the same incorrect way as evaluating the R-Z function at $1$ where it is not defined?' You may as well make something up. I'd go with $f(d) =$ Avocados - I like Avocados :) $\endgroup$ – Zestylemonzi Sep 7 '16 at 18:42
  • $\begingroup$ You probably should say: Let us say we can find sequence $a_n$ such that $$ \sum\limits_{n = 1}^\infty {{a_n} = \frac{{ - 1}}{{12}}} $$ $\endgroup$ – ITA Sep 7 '16 at 18:43
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    $\begingroup$ This is a perfectly sensible question in the context of Zeta function regularization, which is a summability method. There is nothing "mock" or "random" about it. The OP is quite aware of the fact that the series diverges. $\endgroup$ – Robert Israel Sep 7 '16 at 20:02
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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.

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  • $\begingroup$ Interesting. I see that putting this sequence into google tells me that this sequence is the coefficients of the Maclaurin series expansion of $\frac{z}{ln(1+z)}$ called Gregory coefficients. So the answer would be $f(d) = (-1)^dG_{d+1}$ $\endgroup$ – zooby Sep 9 '16 at 0:05

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