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?
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: $$\frac{1}{2},~\frac{-1}{12},~\frac{-12}{12},~\frac{-19}{120},~\frac{-9}{20},\frac{-863}{504},\frac{-1375}{168}$$
My basic idea is, that it might be possible to understand the series $S_k$ as composition of zeta-series where the arguments are negative integers and then assume, that that insertion of the negative integer arguments can be justified by some analytical continuation/zeta-regularization.
For instance beginning for k=2 with $$ S_2 = 1\cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... \tag 1$$ then introducing $$ \begin{align} {} f_2(s) &= {1\cdot 2 \over 1^s} + {2 \cdot 3 \over 2^s} + {3 \cdot 4 \over 3^s} + ... \\ &= 2! \sum _{i=0}^\infty \binom{i+2}{2} { 1\over (1+i)^s}\\ \end{align} \tag 2$$
It is obvious, that there are continuous intervals for s like $ C_k \lt s \lt \infty $ with some fixed constand $C_k$ depending on k where this are convergent expressions. Then I assume, that it is meaningful to set
$$ S_2 = \lim_{s \to 0} f_2(s) \tag 3$$
as a limit-expression or by zeta-regularization or analytical continuation (don't know what the correct expression would be). I think that the general key for the analytical continuation is the said observation, that there is a continuous interval for s where the expression $f_2(s)$ is convergent - because for that s it can freely be decomposed into partial series.
Thus the goal is finding a composition of zeta's at s,s-1,s-2,... valid for continuously and infinitely many $s \gt C_2$ where the series is convergent and then let s=0 .
For the given example and some s (for $s \gt 3 $ this is convergent) we can then rewrite $$ \begin{array} {rll} f_2(s) &= \large {1\cdot 2 \over 1^s} + {2 \cdot 3 \over 2^s} + {3 \cdot 4 \over 3^s} + ... \\ &= \large { 2 \over 1^{s-1}} + { 3 \over 2^{s-1}} + { 4 \over 3^{s-1}} + ... \\ &= \large { 1 \over 1^{s-1}} + { 1 \over 2^{s-1}} + { 1 \over 3^{s-1}} + ... & + \large { 1 \over 1^{s-1}} + { 2 \over 2^{s-1}} + { 3 \over 3^{s-1}} + ... \\ &= \zeta(s-1) & + \large { 1 \over 1^{s-2}} + { 1 \over 2^{s-2}} + { 1 \over 3^{s-2}} + ... \\ &= \zeta(s-1) & + \zeta(s-2) \\ \to S_2 & \underset{\mathcal Z}{=} f_2(0) = \zeta(-1)+\zeta(-2) &= - {1\over12} \end{array} \tag 4\\ $$ where $\mathcal Z$ means the zeta-regularization/ analytical continuation/ limiting-process.
For small k the sums $S_k$ can so be determined by -some tedious- manual pattern-detection, but the pattern-detection provides then also some simple general scheme where the zeta's are to be composed by Stirling numbers first kind which seems to be provable by relative simple compositions of matrices of binomial-coefficients and Stirling numbers.
Let $s_{r,c}$ denote the unsigned Stirling number first kind from the (infinite) array whose indices begin at zero
$$\Large S1 = \small \begin{bmatrix}
1 & . & . & . & . & . \\
0 & 1 & . & . & . & . \\
0 & 1 & 1 & . & . & . \\
0 & 2 & 3 & 1 & . & . \\
0 & 6 & 11 & 6 & 1 & . \\
0 & 24 & 50 & 35 & 10 & 1 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\ddots
\end{bmatrix} \tag 5$$
then the sums in question are expressible in closed forms as $$S_k = \sum_{i=0}^k s_{k,i} \zeta(-i) \tag 6$$
Similarly the $f_k(s)$ - versions are
$$f_k(s) = \sum_{i=0}^k s_{k,i} \zeta(s-i) \tag {6.1} $$
The results for the $S_k$ are then
$$\small \begin{array} {r|r}
k & S_k \\ \hline
0 & -1/2 \\
1 & -1/12 \\
2 & -1/12 \\
3 & -19/120 \\
4 & -9/20 \\
5 & -863/504 \\
6 & -1375/168 \\
7 & -33953/720 \\
\vdots & \vdots
\end{array} \tag 7$$
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).
