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

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 $$\int_0^1\frac{I_1(2\sqrt{-x\ln x})}{\sqrt{-x\ln x}}dx=\sum_{n\ge 0}\frac{(-1)^n}{n!(n+1)!}\int_0^1x^n\ln xdx=\sum_n\frac{\int_0^1u^ne^{-(n+1)u}du}{n!(n+1)!}=\sum_{k\ge 1}\frac{1}{k!k^k}.$$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.
