# 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?

• I want to assign an exact value - I'll edit that in – Matt Majic Mar 30 '16 at 8:56
• Sorry - an exact value by any method -maybe analytic continuation – Matt Majic Mar 30 '16 at 9:08
• I didn't downvote, but it's quite badly phrased. It seems that you want to assign a finite value to a divergent series by some (re)summation technique, but you don't state that in the question -- it sounds as if you want to find the limit of a divergent series. – joriki Mar 30 '16 at 9:34
• OK thanks I re-edited -but of course I'm not asking for a divergent limit, I even included an example which has a value using resummation techniques – Matt Majic Mar 30 '16 at 10:08
• Maybe you can find useful the book of Hardy Divergent Series – alexjo Mar 30 '16 at 10:35

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: $$\frac{1}{2},~\frac{-1}{12},~\frac{-12}{12},~\frac{-19}{120},~\frac{-9}{20},\frac{-863}{504},\frac{-1375}{168}$$

• I like this. Maybe show the actual values for k=0, 1, 2, 3. – marty cohen Mar 31 '16 at 3:47
• Could you please post your method as an answer? – Matt Majic Mar 31 '16 at 23:01
• I deleted my comment saying the magnitudes of our answers are the same. I made a mistake – Matt Majic Apr 1 '16 at 9:46
• I found that using rising factorials gives the same values as your method. I'm not sure what went wrong using falling factorials. – Matt Majic Apr 1 '16 at 11:26
• Just two small bugs: for $k=2$ it should surely be $-1/12$ instead of $-11/12$? And for $k=0$ it should surely be $-1/2$ instead of $1/2$ because this should equal $\zeta(0)$? – Gottfried Helms Apr 1 '16 at 11:29

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 $$\small \begin{array} {r|r} k & S_k \\ \hline 0 & -1/2 \\ 1 & -1/12 \\ 2 & -1/12 \\ 3 & -19/120 \\ 4 & -9/20 \\ 5 & -863/504 \\ 6 & -1375/168 \\ 7 & -33953/720 \\ \vdots & \vdots \end{array} \tag 7$$

For a further justification of this method I looked with the same scheme at the alternating series $A_k$ istead of the original series $S_k$. Also using the alternating zeta-series $\eta()$ instead gives $$A_k = \sum_{i=0}^k s_{k,i} \eta(-i) \tag 8$$ and identical results pop up as when I do the sum of the explicite alternating series using Euler-summation. Here the results are $$\small \begin{array} {r|r} k & A_k \\ \hline 0 & 1/2 \\ 1 & 1/4 \\ 2 & 1/4 \\ 3 & 3/8 \\ 4 & 3/4 \\ 5 & 15/8 \\ 6 & 45/8 \\ 7 & 315/16 \\ \vdots & \vdots \end{array} \tag 9$$
Final remark: it might be interesting, that the numerators and denominators of the $S_k$ are in the OEIS, see numerators and denominators (the numerators are given with alternating signs) and there is a whole bunch of furtherly interesting links and references!

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).

• How did you get the first equality? And the final result should be $k!/(-2)^{k+1}$ right? – Matt Majic Mar 31 '16 at 22:57
• I think that solution is not optimal. For instance, the same scheme should be applicable for the alternating version of the series, and using the $\eta()$ - function instead of the $\zeta()$-function. The advantage of such an application would be, that for the evaluation of the alternating series we have concurring evaluation-methods like Abel and Euler-summation. I've checked that scheme which I used in my answer (applied to the $\eta()$-function) and that is compatible with the results for the alternating series taken by Euler-summation. – Gottfried Helms Apr 1 '16 at 8:42
• @GottfriedHelms Wrong. The elementary Ramanujan summation gives the same result for alternating series as those methods because it is an extension of the Abel summation. – user299632 Apr 1 '16 at 13:15
• What is "wrong"? That I've tested the alternating series? That I found the results comparing two methods compatible? That the alternating-series testing should be compatible with using the eta-values instead of zeta-values? – Gottfried Helms Apr 1 '16 at 13:21
• @MattMajic Thus for $k=0$, you have $\sum_{n=0}^\infty 1=1+\sum_{n=1}^\infty 1$=1+\zeta(0)=1/2$, not$-1/2\$. – user299632 Apr 2 '16 at 13:10