Probability of convergence Given a number $x\in (0,1)$ with decimal representation $0.a_1a_2a_3...$. Let $f(x)= 1$ iff the sequence 
$\sum_{i=1}^{n} {\frac{a_i}{n}}$ converges. Takes $X$ to be uniform$(0,1)$, what is the expected value of $f(X)$? Also, what can we say about the set of $x$ such that the sequence diverges?
Edit: I'm fairly sure the answer to the first question is 1. I'm mostly interested now in whether the set of divergent x has any nice properties. 

Juse to talk about the (sub)set of diverging $x$ here.
The sequence $c_n = \frac1n\sum_{i=1}^na_i$ is known as the Cesàro mean. One way to make the mean diverge with a bounded sequence is to make it slowly diverging, e.g. let
$$a_i=(-1)^{\lfloor \log_2 i \rfloor} + 1 \implies x=0.2\underbrace{00}_2\underbrace{2222}_4\underbrace{00000000}_8\underbrace{2222222222222222}_{16}\dots,$$
then we could find two subsequences with different limits
$$c_{2^k-1} = \begin{cases} \frac23 & \text{$k$ is even}\\ \to \frac43 & \text{$k$ is odd}, \end{cases}$$
i.e. $c$ diverges and $f(x)=0$.
Any diverging sequence $b_i$ can be converted to supersequence $a_i=b_{T(i)}$ with $T(i)$ being logarithmic or slower which the Cesàro mean diverges. 
