Divergent sum of factorials Is it possible to get an exact value of the sum (using divergent series summation methods)
$$ \sum_{n=0}^\infty~ \frac{(n+k)!}{n!} \quad?$$
where $k$ is a positive integer.
The only other divergent sum of factorials I have seen is $\sum_{n=0}^\infty(-1)^nn!$.
Does anyone know any useful techniques or references?
 A: I don't know how to check this, but here might be one possible approach using the Riemann zeta function:
You can write the factorials as a rising factorial and express this as a sum of powers using the unsigned Stirling numbers of the first kind:
\begin{align}
   \sum_{n=0}^\infty~\frac{(n+k)!}{n!} &= \sum_{n=0}^\infty~(n+1)^{(k)} \\
&= \sum_{n=0}^\infty~\sum_{p=0}^k~ {k\brack p}(n+1)^p \\
&= \sum_{p=0}^k~ {k\brack p} \sum_{n=0}^\infty~ (n+1)^p \\
&=-\sum_{p=0}^k~ {k\brack p}\frac{B_{p+1}}{p+1}
\end{align}
where $B_p$ are the Bernoulli numbers (you must use $B_1=1/2$). The last line is not a real equality. This is the same as Gottfred's answer.
The first few values are:
$$\frac{1}{2},~\frac{-1}{12},~\frac{-12}{12},~\frac{-19}{120},~\frac{-9}{20},\frac{-863}{504},\frac{-1375}{168}$$
A: We have
$$
\sum_{n=0}^\infty \frac{(n+k)!}{n!}x^n=\frac{d^k}{dx^k}\sum_{n=0}^\infty x^{n+k}=\frac{d^k}{dx^k}\frac{x^k}{1-x}=\frac{k!}{(1-x)^{k+1}}
$$
for $|x|<1$ and may use the elementary Ramanujan summation (definition) (or simply linearity) and obtain your result, which corresponds to $k!(\sum_{n=0}^\infty 1)^{k+1}$
as a Cauchy product of series (here and here) (not $k!(-1/2)^{k+1}$).
Example $k=0$. We have
$$
\sum_{n=0}^\infty x^n=\frac{1}{1-x}=\frac{x}{1-x}+1~.
$$
As $x/(1-x)=x+x^2+x^3+\cdots$ for $|x|<1$, corresponding to $1+1+1+\cdots=-1/2=\zeta(0)=\sum_{n=1}^\infty 1$, we obtain $\sum_{n=0}^\infty 1=1-1/2=1/2$.
Example $k=1$. We have
$$
\sum_{n=0}^\infty (n+1)x^n=\sum_{n=0}^\infty nx^n + \sum_{n=0}^\infty x^n=\frac{x}{(1-x)^2}+\frac{1}{1-x}=\frac{1}{(1-x)^2}~.
$$
As $x/(1-x)^2=x+2x^2+3x^3+\cdots$ for $|x|<1$, corresponding to $1+2+3+\cdots=-1/12=\zeta(-1)=\sum_{n=1}^\infty n$, we obtain
$\sum_{n=0}^\infty (n+1)=-1/12+1/2=5/12$.
Example $k=2$. We have
$$
\sum_{n=0}^\infty (n+1)(n+2)x^n=\frac{x+x^2}{(1-x)^3}+\frac{3x}{(1-x)^2}+ \frac{2}{1-x}=\frac{2}{(1-x)^3}~.
$$
As $(x+x^2)/(1-x)^3=x+4x^2+9x^3+\cdots$ for $|x|<1$, corresponding to $1+4+9+\cdots=0=\zeta(-2)=\sum_{n=1}^\infty n^2$, we obtain $\sum_{n=0}^\infty (n+1)(n+2)=0+3\times(-1/12)+2\times(1/2)=3/4$.
The elementary Ramanujan summation of a series is consistent with the analytic continuation of Dirichlet series (here, here and here).
For the example $k=2$, using the venerable method of analytic continuation of Dirichlet series to sum the series $\sum_{n=0}^\infty (n+1)(n+2)=2+\sum_{n=1}^\infty (n^2+3n+2)$, corresponding to
$$
F(s)=2+\sum_{n=1}^\infty(n^2+3n+2)n^{-s}=2+\zeta(s-2)+3\zeta(s-1)+2\zeta(s)~,
$$
we obtain $F(0)=2+\zeta(-2)+3\zeta(-1)+2\zeta(0)=2+0+3\times(-1/12)+2\times(-1/2)=3/4$, in agreement with the elementary Ramanujan summation.
It is well-known that, in general, the sum of a divergent series and the sum of the shifted series are different (here and here).
