Does the series $ \sum_k^\infty \frac{k!}{k^k}$ converge to an irrational number and does it have any significance or applications? I know that the series 
$$ \sum_k^\infty \frac{k!}{k^k} $$
converges by the ratio test.
The sum calculated by wolfram alpha is $~1.87985386217525853348630614507096$ which seems pretty irrational to me.
But is it in any way significant? It there a proof that it is an irrational number? Are there any applications of this series?

I completely forgot about the Stirling's approximation for the factorial.
Since for $k  \rightarrow +\infty$ the factorial approaches
$$
k! \sim  \sqrt{2 \pi k} \frac{k^k}{e^k}
$$
So it seems obvious that the series give irrational number.
 A: It may be relevant to notice that:
$$\begin{eqnarray*}\sum_{k\geq 1}\frac{k!}{k^k}=\int_{0}^{+\infty}\sum_{k\geq 1}\frac{x^k}{k^k}e^{-x}\,dx&=&\int_{0}^{+\infty}\sum_{k\geq 1}(-1)^{k-1} \frac{x^k}{(k-1)!} e^{-x}\int_{0}^{1}(y\log y)^{k-1}\,dy\,dx\\&=&\int_{0}^{+\infty}\int_{0}^{1}xe^{-x}\cdot y^{-xy}\,dy\,dx\\&=&\int_{0}^{1}\frac{dy}{(1+y\log(y))^2}\\&=&\int_{0}^{+\infty}\frac{dz}{e^{z}\left(1-z e^{-z}\right)^2}.\end{eqnarray*}$$
Now the classical approach for showing that some constant is an irrational number is to provide some accurate rational approximation, too accurate to be approximations for a rational number.  The last integral is not the most manageable I have met in my life, but Laguerre polynomials may provide a good way to go.
On the other hand, only the order of growth of the coefficients of a series converging to $\alpha$ tells us nothing about the rationality or irrationality of $\alpha$. Consider, for instance:
$$\sum_{n\geq 1}\frac{1}{n^3}=\zeta(3)\not\in\mathbb{Q},\qquad \sum_{n\geq 1}\frac{F_n}{19^n}=\frac{19}{341}\in\mathbb{Q}.$$
