Has someone seen a discussion of the (divergent) summation of $\sum\limits_{k=0}^\infty (-1)^k (k!)^2 $?

Hopefully, this answer is not too late. Neither of the previous answers seemed to be entirely satisfactory, and I hope this might be useful.

Consider the Mellin transform identity

$$\int_0^\infty u^k (2\,K_0(2\sqrt u))\,\mathrm du=(k!)^2$$

where $K_0(z)$ is the modified Bessel function of the second kind.

Now, recall the fundamental (formal) relationship for the Stieltjes moment problem:

$$\int_0^\infty\frac{\mathrm d\,\eta(t)}{z+t}=\sum_{k=0}^\infty \frac{(-1)^k \eta_k}{z^{k+1}}$$

where $\eta$ is some positive measure on $(0,\infty)$ and $\eta_k=\int_0^\infty t^k \mathrm d\,\eta(t)$ are the corresponding Stieltjes moments. Making the appropriate replacements, we have

$$2\int_0^\infty\frac{K_0(2\sqrt t)\,\mathrm dt}{z+t}=\sum_{k=0}^\infty \frac{(-1)^k (k!)^2}{z^{k+1}}$$

In particular, for $z=1$, we have the formal identity

$$2\int_0^\infty\frac{K_0(2\sqrt t)\,\mathrm dt}{t+1}=\sum_{k=0}^\infty (-1)^k (k!)^2$$

Mathematica tells me that the integral can be expressed in terms of the Meijer $G$-function:

$$2\int_0^\infty\frac{K_0(2\sqrt t)\,\mathrm dt}{z+t}=G_{1,3}^{3,1}\left(z\mid{{0}\atop{0,0,0}}\right)$$

Here is a Mathematica verification that the function given above has the appropriate asymptotic expansion:

Series[MeijerG[{{0}, {}}, {{0, 0, 0}, {}}, z], {z, ∞, 10}]
   1/z - (1/z)^2 + 4/z^3 - 36/z^4 + 576/z^5 - 14400/z^6 + 518400/z^7 -
   25401600/z^8 + 1625702400/z^9 - 131681894400/z^10 + O[1/z]^11

% - Sum[((-1)^k (k!)^2)/z^(k + 1), {k, 0, 10}]
   O[1/z]^11

Either numerical evaluation of the integral or the Meijer $G$ expression yields the result

$$G_{1,3}^{3,1}\left(1\mid{{0}\atop{0,0,0}}\right)\approx0.668091326377777765433014987501$$

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I have been asked how I came to the Mellin transform at the beginning of this post. Here's how I did it.

I started with the inverse problem; that is, I needed the inverse Mellin transform of $\Gamma(s+1)^2$:

$$\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(1+s)^2 t^{-s}\,\mathrm ds$$

This can be recognized as a particular case of the Meijer $G$-function, if you compare this with the definition. In particular,

$$\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(1+s)^2 t^{-s}\,\mathrm ds=G_{0,2}^{2,0}\left(t\mid{{—}\atop{1,1}}\right)$$

As it turns out, this case was listed in the Wolfram Functions site:

$$G_{0,2}^{2,0}\left(t\mid{{—}\atop{1,1}}\right)=2t\,K_0(2\sqrt t)$$

Inverting this relationship gives the Mellin transform identity for the square of the factorial.

is borel summable...provided the integral

$$ S= \int_{0}^{\infty}dt \exp(-t) \int_{0}^{\infty} \frac{e^{-u}}{1+ut} $$ exists

because $$ \sum_{n=0}^{\infty}(-1)^{n}u^{n} $$ is BOREL SUMMABLE to the exponential integral

$$ \int_{0}^{\infty}dt \frac{e^{-t}}{1+xt} =E(x) $$

or you can use Mellin convolution $ M(e^{-t}*e^{-t})= (s)!^{2} $ defined by

$$ (f * g)(x) = \int_{0}^{\infty} f(t)g(t^{-1}x)dt. $$

The use of the "Eulerian-summation" procedure seems to arrive at the same value which J.M. has provided. It seems to be a weak procedure; doubling the number of involved terms gives only two more digits precision; here is a table (correct digits before the centered-dots): $$ \small \begin{array} {r|l} \text{n terms} & \text{partial sums} \\ \hline 16 & 0.668\cdot 10011598092309157 \\ 32 & 0.66809\cdot 094424944193264 \\ 63 & 0.66809132\cdot 812814354929 \\ 64 & 0.66809132\cdot 829512992943 \\ 128 & 0.66809132637\cdot 441027814 \\ \hline JM & 0.668091326377777765433014987501 \end{array} $$

This "Eulerian summation" involves the computation of the integral $ \displaystyle f(x) = \int_0^\infty {e^{-u} \over 1 - ux} du $ which is asymptotic to $ f(x) \sim 1+1x+2!x^2+3!x^3+4!x^4 + \ldots$ and its derivatives (with respect to x, not to u(!)) at negative integers, which can advantageously be expressed by some polynomials (and OEIS) in $ {1 \over x^k}$ cofactored with the original computation of $f(x)$ only. However, the numerical integration is rather time consuming, so the whole procedure seems to be somehow inefficient and is thus finally practically uninteresting.
I'll "accept" now @J.M.'s answer because that gave the first trustworthy result.

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Why it might still be a principally interesting procedure should then be discussed elsewhere.