Infinite sum of random numbers Let $x_n$, with $n=1,2,\ldots$, be uniformly distributed random variables in $(0,1)$
What is the expected value and probability distribution of the sum  
$$\sum_{n=1}^\infty x_n^{n^n}$$
 A: Elaborating on Steven's answer.
Since $\sum\nolimits_{n = 1}^\infty  {{\rm E}[X_n^{n^n } ]}  < \infty $ and $\sum\nolimits_{n = 1}^\infty  {{\rm Var}[X_n^{n^n } ]}  < \infty $, the series $\sum\nolimits_{n = 1}^\infty  {X_n^{n^n } } $, assuming that the $X_n$ are independent, converges with probability $1$ (see Theorem 1 on page 130 of the book Probability theory: collection of problems). Call the limit $X$. Then, by the monotone convergence theorem, 
$$
{\rm E}[X] = \mathop {\lim }\limits_{n \to \infty } {\rm E}\bigg[\sum\limits_{k = 1}^n {X_k^{k^k } } \bigg] = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{\rm E}[X_k^{k^k } ]}  = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{k^k  + 1}}}  = \sum\limits_{n = 1}^\infty  {\frac{1}{{n^n  + 1}}} .
$$ 
EDIT: The fact that the series $\sum\nolimits_{n = 1}^\infty  {X_n^{n^n } } $ converges with probability $1$ follows from the monotone convergence theorem (MCT), and we don't have to assume that the $X_n$ are independent (it suffices that they are positive). Indeed, by MCT, the limit $X:=\sum\nolimits_{n = 1}^\infty  {X_n^{n^n } }$ must be almost surely finite, since the integral $\int_\Omega  {XdP} \,(= \sum\nolimits_{n = 1}^\infty  {\frac{1}{{n^n  + 1}}})$ is finite. 
A: Your answer almost certainly has no closed form, but it's easy enough to write the EV as a sum; just use the linearity of expectation, and use the defining integral to calculate the expectation of each of the individual terms.
