# Infinite Sums of reciprocal of factorial times power

I came across this crazy sum, and I have no idea how to tackle it

$$\sum_{k=1}^{\infty}\frac{1}{k!k^k}.$$

I tried to approximate it using wolfram alpha widget,

and got something like $$1.13134$$.

However, I'd like to get an exact solution.

EDIT: i was thinking that maybe we could play around with the ceiling function a little bit. Consider that:

$$\sum_{k=1}^{\infty}\frac{1}{k!k^k}.$$

is actually the area under the curve

$$f(k)=\frac{1}{\left\lceil{k^k}\right\rceil}\frac{1}{\left\lceil{k!}\right\rceil}$$

from $$0$$ to $$\infty$$. More formally, this can be expressed as: $$I=\int_0^\infty \frac{1}{\left\lceil{k^k}\right\rceil}\frac{1}{\left\lceil{k!}\right\rceil} dk=\sum_{k=1}^{\infty}\frac{1}{k!k^k}.$$

If we multiply $$I$$ by: $$L=\int_0^\infty{\left\lceil{k^k}\right\rceil}{\left\lceil{k!}\right\rceil} e^{-k}dk.$$

We get that:

$$I L=\int_0^\infty e^{-k} dk=1.$$

Hence:

$$\sum_{k=1}^{\infty}\frac{1}{k!k^k}\int_0^\infty{\left\lceil{k^k}\right\rceil}{\left\lceil{k!}\right\rceil} e^{-k}dk=1,$$

so if we could somehow evaluate $$L$$, then it might be possible to calculate the sum.

Unfortunately, I'm not sure if this reasoning works.

• Exact solution, seems impossible with Earth math. – Mariusz Iwaniuk Apr 13 at 18:42
• ${\rm arcsinh} \left(3/5\,{\frac {{2}^{2/3}{3}^{3/5}}{ \left( \ln \left( 3 \right) \right) ^{3}}}\right)$: Exact solution,but with only 9 correct digits. – Mariusz Iwaniuk Apr 13 at 19:07
• $\displaystyle{1 \over k!\, k^{k}} = \left[z^{k}\right]\mathrm{e}^{z/k}$. – Felix Marin Apr 13 at 19:38

In a variant on the famous sophomore's dream calculation, note first that, since in terms of modified Bessel functions $$\sum_{n\ge 0}\frac{y^n}{n!(n+1)!}=\frac{I_1(2\sqrt{y})}{\sqrt{y}}$$, the substitution $$u=-\ln x$$ obtains $$\int_0^1\frac{I_1(2\sqrt{-x\ln x})}{\sqrt{-x\ln x}}dx=\sum_{n\ge 0}\frac{(-1)^n}{n!(n+1)!}\int_0^1x^n\ln xdx=\sum_n\frac{\int_0^1u^ne^{-(n+1)u}du}{n!(n+1)!}=\sum_{k\ge 1}\frac{1}{k!k^k}.$$Now we just need to evaluate that integral... which seems equally impossible. I also had no look identifying the value with the inverse symbolic calculator.