So I came across this exposition of a paper by Euler here 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:

  1. 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.

  2. 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.

  3. 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.


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", is this still the best book to go to?

  • 2
    $\begingroup$ This very example is discussed on Wikipedia's page on Borel summation. $\endgroup$ – Semiclassical Jun 19 '16 at 21:15
  • $\begingroup$ I like the book of K. Knopp even more, because he gives also a lot of exercises. (Of course G.H.Hardy's monography is the final reference) When I started to look at divergent series I found really plenty online-articles about them, a very fascinating subject, indeed! $\endgroup$ – Gottfried Helms Jun 20 '16 at 4:15
  • $\begingroup$ You do not name "this exposition". Perhaps you mean eulerarchive.maa.org/hedi/HEDI-2006-06.pdf $\endgroup$ – Noam D. Elkies Jun 20 '16 at 6:00
  • $\begingroup$ @NoamD.Elkies Yes, that is the correct article. I linked it in the first sentence (at "here") but perhaps I should have been more explicit. $\endgroup$ – Asvin Jun 20 '16 at 7:49
  • $\begingroup$ Sorry, I didn't notice the color change that indicates a hyperlink. $\endgroup$ – Noam D. Elkies Jun 20 '16 at 13:22

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


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