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.

`helms (at) uni-kassel.de`

$\endgroup$`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 $$ 0.70416996043747446070 \\ 0.70416996043747446035 \\ 0.70416996043747446017 \\ 0.70416996043747446011 \\$$ $\endgroup$1more comment