The divergent sum of alternating factorials So I came across this exposition of a paper by Euler1  where Euler is trying to sum the divergent sum:
$$s = 1 - 1 + 2! - 3! + 4! \dots = \sum_{k\geq 0}(-1)^k k!.$$
There are a couple of questions at the end. However, I would like to go through some exposition first:
This sum is certainly divergent but this of course does not stop Euler. He goes through various manipulations and always ends up at the same value of $s \sim 0.5963$. Briefly, the various methods are as follows:

*

*Let $s = \sum_{k\geq 1}(-1)^ka_k$ with $a_k > 0$. He defines:
$$b_l = \sum_{0\leq k\leq l}(-1)^{l-k}\binom{l}{k}a_{k+1}$$
and then shows that:
$$s = \sum_{l\geq 1}\frac{b_l}{2^l}.$$
We can apply this to our $s$ repeatedly and find an approximate value.


*It is mentioned that he finds diverging series for $1/A$ and $\log A$ and using similar methods finds the same approximate value. He also manages to find continued fraction expansions for $A$ and $1/A$ and approximates them to the same value.


*Finally, the linked paper goes through the following fascination derivation. Define:
$$s(x) = \sum_{k\geq 0}(-1)^kk!x^{k+1}.$$
Then:
$$\frac{ds}{dx} = \frac{x-s}{x^2}$$
and Euler solves this differential equation to get the integral representation:
$$s(x) = e^{1/x}\int_0^x\frac{e^{-1/t}}{t}dt$$
and evaluating at $x=1$ ends up with the same approximation.
Questions:
There are a few obvious questions here. There seems to be a strong internal consistency to this series. Why? Usually, I would suspect that there is some analytic continuation lurking in the background. Is this true?
Second, what is the exact value that we are approximating?
Third, is there any book/article with more on divergent series like these. This is all incredibly fascinating and I would love to learn more. The article references Hardy's "Divergent Series"2 , is this still the best book to go to?

1 Sandifer, C. Edward, How Euler did it, The MAA Tercentenary Euler Celebration 3. Washington, DC: Mathematical Association of America (MAA) (ISBN 978-0-88385-563-8/hbk). xiv, 235 p. (2007). ZBL1141.01009.  Link on Euler Archive
2 Hardy, G. H., Divergent series, Oxford: At the Clarendon Press (Geoffrey Cumberlege) XIV, 396 S. (1949). ZBL0032.05801. Link on Internet Archive
 A: I think, that an "analytical continuation" is impossible for this series because the powerseries in $x$ has zero-convergence radius and thus the method of recentering the series to extend its evaluatable range  step-by-step cannot be exploited here.

Because you said you like the problem of divergent series - here some (amateurish, but I think: really nice) approach of myself, which vaguely resembles the Borel-summation:
Once I've found a nice method to sum this series and to assign a meaningful value to it (it's the same value found by Euler's method). I'm using a "matrix-method", but in some reverse order...
First I consider the infinite sized matrix of Eulerian numbers
    matrix E                  rowsums
  1   .   .   .  .  .  ...     1 = 0! 
  1   0   .   .  .  .          1 = 1!
  1   1   0   .  .  .          2 = 2!
  1   4   1   0  .  .          6 = 3!
  1  11  11   1  0  .         24 = 4!
  1  26  66  26  1  0   ...  120 = 5!
  ...  ...  ...  ... ...

To compute the rowsums in a matrix-formula let us introduce the column-vector of only ones $U=[1,1,1,1,...] ^\tau $ . The row-sums (the factorials per row) can then be expressed as $E \cdot U = F $ where $F=[0!,1!,2!,3!,...] ^\tau$ . Of course, if we rescale the rows by just that factorials we get
 $$ diag(F)^{-1} \cdot ( E \cdot U )= U $$ and if we have a series and write its terms in a rowvector $A=[a_0,a_1,a_2,...] $ we can express the summing of the series by $A \cdot U = s $ also by $$A \cdot \left( diag(F)^{-1} E U \right) $$ and even more, if the left dot-product has convergent summations we can even associate the matrix-product differently by
$$ s = \left( (A \cdot diag(F)^{-1}) \cdot E \right) \cdot U $$
Now we see, that, given $A=[0!,-1!,2!,-3!,...]$ multiplied by the inverse factorials we have $ A \cdot diag(F)^{-1} = [1,-1,1,-1,1,-1,...] $ Let's denote this last vector by $W$, then the dot-products with the Eulerian numbers $W \cdot E$ give (columnwise) series expressions like the one with the first column:
$$ w_0 =[1,-1,1,-1,1,-1,...] \cdot [1,1,1,1,1,...] = \sum_{k=0}^\infty (-1)^k =  1/2 $$
which is not a convergent series, but an alternating geometric series to which we assign the value of the closed fractional form $ {1 \over 1+1}=\frac 12$ .
With the following columns this is not so obvious. Fortunately we have an analytic description of that matrix-entries along the columns: they can be understood as sums of terms of geometric series and their derivatives again, for instance $ [0,1,4,11,26,...]=[1-1,2-2,2^2-3,2^3-4,2^4-5,2^5-6,...]$ and using this for the dotproduct with $W$ gives the sum for the alternating geometric series with quotient 2 minus the first derivative of the alternating geometric series qith quotient 1, which is explicitely 
$$ w_1 = W \cdot E_1 = \sum_{k=0}^\infty (-2)^k - \sum_{k=0}^\infty \left((-1)^k \cdot (k+1)\right) = \small {{1 \over 1+2} - \left(- \left({1 \over 1+1}\right)^2 + {1 \over 1+1}\right) = \frac 1{12}}
$$
In a similar way we get for the next colum $w_2 = \frac 1{72}$ and so on. The final evaluation reads then
$$ \left(W \cdot E \right) \cdot U = [{1\over2},{1\over 12},{1\over 72},{1\over 1080},...]\cdot U \approx   0.596347362323...
$$
This is just an exemplaric illustration. Analytically it is better to introduce a continuous argument $x$ into the system, such that the vector $A$ becomes A(x)=$[1,1!x,2!x^2,3!x^3,...]$ and then evaluate that this holds for the indeterminate $x$ and then also for $A(x)_{|x=-1}=A(-1)$ by the identities for the alternating geometric series and their derivatives at $x$ .                
A more complete explanation is this essay at my math-pages
A: From an engineering point of view, the concept is to apply to the "digital signal"
represented by a sequence or sequence of partial sums, 
a Low-pass filtering so as to level-out 
the  transient response however
oscillating  it might be, and even if  with diverging amplitude.
What is retained is the $f = 0$ ( $f \to 0$) component.
So in the Grandi's series, with partial sums $1,0,1,0,\cdots$, the steady component = mean value
is clearly $1/2$.
Also it is evident that in the Cesàro summation 
the average of the sequence of the partial sums is just the $a_0$ coefficient of the corresponding
Fourier series, with $T \to \infty$ and thus with the Fourier series tending to the transform.
To focalize the steady component we need to compress the diverging (if it is) amplitude of the oscillation, 
by applying a filter with a stiffness of an order high enough and possibly having a mathematical formulation
that pretty matches and annihilates the divergency of the "noise".   
In conclusion a "filter" is conceptually the same as a sequence transformation
and related summation approaches.
And same  as the characteristics of the signal indicate the type and parameters of the filter which is better to adopt, so it is for the 
sequence transformation.
