Value of divergent sum $\sum_{n=0}^\infty (-1)^n n^n$

Question

I'm hoping to find a reasonable value to assign to the divergent series $\sum_{n=0}^\infty (-1)^n n^n$ and $\sum_{n=0}^\infty (-1)^n (xn)^n$. For the first one, I have obtained something around 0.71, but I'm very unsure if this is correct. For the second series, I get a graph that looks about like this: .

What sort of methods are powerful enough to sum this series? I think Borel summation is too weak, since this grows faster than any $\left(\alpha n\right)!$.

Any help would be appreciated!

EDIT: I obtained my values from a series of questionably valid approximations, but here is another method that seems to give a similar value.

The idea is that illegally swapping summation signs, and then continuing the inner summation past its regular convergent range allows one to assign a value to a divergent series.

Starting with $$\sum_{n=0}^\infty (-1)^n n^n$$ We want to get two summations in order to be able to swap, so we expand into $$\sum_{n=0}^\infty (-1)^n e^{n\ln(n)} = \sum_{n=0}^\infty (-1)^n \sum_{k=0}^\infty \frac{\left(n \ln(n)\right)^k}{k!}$$ Simplifying a bit more, we get $$\sum_{n=0}^\infty (-1)^n \sum_{k=0}^\infty \frac{n^k \ln(n)^k}{k!} =\sum_{n=0}^\infty (-1)^n \sum_{k=0}^\infty \frac{e^{\ln(n)k} \ln(n)^k}{k!}$$ Swapping the summations, we get $$\sum_{k=0}^\infty \frac{1}{k!} \sum_{n=0}^\infty (-1)^n e^{\ln(n)k} \ln(n)^k = \sum_{k=0}^\infty \frac{1}{k!} \sum_{n=0}^\infty \frac{d^k}{dk^k} (-1)^n n^k $$ (Note: I'm using $\frac{d^k}{dk^k}$ to represent taking the derivative with respect to k, k times). Now, we can continue the inner summation by doing $$\sum_{k=0}^\infty \frac{1}{k!} \frac{d^k}{dk^k}\sum_{n=0}^\infty (-1)^n n^k = \sum_{k=0}^\infty \frac{1}{k!} \frac{d^k}{dk^k}(1-\eta(-k))$$ To finally get $$1/2 - (\eta'(-1) - \frac{\eta''(-2)}{2!} + \frac{\eta'''(-3)}{3!} - \dots)$$ The sum of the first few terms seems to be around .71, which agrees fairly closely with the other method.

EDIT2: It looks like most methods converge onto about the same shape eventually. The green line is the asymptotic expansion, meaning I only took the first 5 terms in the power series, and left the rest out. Usually, this asymptotic expansion converges to the right function within a very small radius. The black function is using something like FusRoDah's method. The orange graph is using the non-rigourous series of approximations.

This next pictures shows the functions over larger intervals--the purple function is using that eta method I outlined above.

To elaborate more about how I used FusRoDah's method, I started with $BA(t)=\sum_{k=0}^\infty \frac {(-kt)^k} {k!} = 1+\sum_{k=1}^\infty \frac {(-kt)^k} {k!}$. Then, I added in the approximation to get $$\operatorname{BA}(t) = 1+\sum_{k=1}^\infty \frac {(-kt)^k} {k!} + \frac {(-et)^k} {\sqrt{2\pi k}}-\frac {(-et)^k} {\sqrt{2\pi k}} = 1+\sum_{k=1}^\infty\frac {(-et)^k} {\sqrt{2\pi k}}+ \sum_{k=1}^\infty \frac {(-kt)^k} {k!} -\frac {(-et)^k} {\sqrt{2\pi k}}.$$ The first part of the sum can be written as an integral of the polylogirthmn with some other terms, but we are still left with the $\int_{0}^{\infty}e^{-t}\sum_{k=1}^{\infty}\left(\frac{\left(-ktz\right)^{k}}{k!}-\frac{\left(-etz\right)^{k}}{\sqrt{2\pi k}}\right)dt$ term. Since this doesn't converge on its own, we can approximate it with $$\int_{0}^{A}e^{-t}\sum_{k=1}^{B}\left(\frac{\left(-ktz\right)^{k}}{k!}-\frac{\left(-etz\right)^{k}}{\sqrt{2\pi k}}\right)dt.$$ Increasing $A$ and $B$ increases the accuracy, but decreases the range of convergence. I used $A = 2.5$, $B = 30$ for the first graph.

What remains to be done then is either find a more accurate approxation, such that $\sum_k \frac{(-kt)^k}{k!} - f_t(k)$ converges for all values of t, or to find a way to continue the difference between the approximation and the original series.

Answer 1

The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x))}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ as it can be seen in the following Pari GP v2.14.0 lines

although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$

Answer 2

The sum we consider is $\sum_{n=1}^\infty (-nz)^n$. (The $n=0$ term causes problems later on.) The Borel transform of this sum is $BA(t)=\sum_k \frac {(-kt)^k} {k!}$. I don't see how we could make progress with this sum, but using Stirling's formula, it should be almost equal to $\sum_k \frac {(-et)^k} {\sqrt{2\pi k}}$. This series only converges for $|t|<1/e$, but can be analytically continued to the so-called polylogarithm $(2\pi)^{-1/2} Li_{1/2}(-et)$. This can the be integrated in the usual way, in which one proceeds with Borel summation. In the case $z=1$, a quick WolframAlpha computation gives $-0.33645$, which should be close to the "value" of your sum.