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
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
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