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

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.

• Could you elaborate on how you obtained the 0.71 value? Sep 6, 2021 at 20:29
• @MaxMuller I obtained .71 from a series of very unrigorous approximations, that is why I'm hoping someone can provide a more reasonable solution so I can figure out which parts of my approximations are reasonble and which parts are not. However, I'm thinking of another method to solve this that I will edit in, but its similarly unrigorous Sep 6, 2021 at 20:35
• I think that it follows from Stirling's approximation that $n!<n^n e^n$, so Borel summation should be fine. However, expecting some kind of "nice" formula for the sum is probably too optimistic. Sep 6, 2021 at 21:05
• For some years I found with my methods $w \approx 1- 0.29632 = 0.70368$ I used as well the double-summation-formula which occurs by expanding the implicite exponential-series, and also applied a (Carleman-)matrix approach. I did never improve that purely explorative draft computation and discussion from 2007 (go.helms-net.de/math/tetdocs/Tetra_Etaseries.pdf) , maybe there's something interesting in it for you. If there is something in it, I'd propose to use email-conversation instead of littering your question here. helms (at) uni-kassel.de Sep 9, 2021 at 9:56
• The sum-formula beginning with $\sum \frac1{k!} \sum \cdots$ can be evaluated using Pari/GP's sumalt()-function for the inner alternating sums followed by a simple Euler-summation of that intermediate results with small order. Using $64$ terms for the outer sum I have this Pari/GP-formulation: tmp=Mat(vectorv(64,k,1/k!*sumalt(n=1,(-1)^n*n^k*log(n)^k))) and for display ESum(1.3,64)*tmp + 0.5*mV(1,64) giving the $60...64$'th approximations $$0.70416996043747446070 \\ 0.70416996043747446035 \\ 0.70416996043747446017 \\ 0.70416996043747446011 \\$$ Sep 13, 2021 at 11:58

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 $$g(x) = W_{0}(x)/(1 + W_{0}(x))^3$$

• Wow! This is truly incredible-- thank you for this answer! Sep 7, 2021 at 3:17
• I should add to this wonderful answer, that $$\int_0^\infty \frac{e^{-x}}{1+W_0(x)}dx = \int_1^\infty \frac{1}{x^x}$$ Which makes this answer answer very similar to form to the sophmore's dream Sep 7, 2021 at 16:16
• I thought about this method, but did not know how to extend $\sum _{k=0}^{\infty } \frac{(-1)^k k^k t^k}{k!}$ analytically. How to prove that this sum is $1/(1+W_0(t))$? Sep 8, 2021 at 14:01
• Iosif, the derivative relationship 1 - t*dW0(t)/dt =1/(1+W0(t)) makes the trick. This sum is an excellent probe test for summation methods acting on alternating divergent series. Sep 9, 2021 at 0:01
• @JorgeZuniga : Thank you for your response. Sep 9, 2021 at 23:39

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.

• I think your value that you assign to the sum agrees with mine, I'll post more details after I finish working it out Sep 6, 2021 at 22:28