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

Question

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.

Answer 1

I'm afraid I have to concur with @MariuszIwaniuk. I'll summarise what I found.

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 $$\int_0^1\frac{I_1(2\sqrt{-x\ln x})}{\sqrt{-x\ln x}}dx=\sum_{n\ge 0}\frac{(-1)^n}{n!(n+1)!}\int_0^1(x\ln x)^ndx=\sum_n\frac{\int_0^1u^ne^{-(n+1)u}du}{n!(n+1)!}=\sum_{k\ge 1}\frac{1}{k!k^k}.$$Now we just need to evaluate that integral... which seems equally impossible. I also had no luck identifying the value with the inverse symbolic calculator.