$$F\\,(s) = 1/1^s + 0/2^s + 1/3^s + 1/4^s + 0/5^s + 1/6^s + 0/7^s + 1/8^s + ... $$
The coefficients of this Dirichlet series are the base-2 digits of $\sqrt2$.
Does it have an analytic continuation into the critical strip?
Assuming so, can anything be said about the locations of its poles in this region? Are there any? Are they all on the critical line?
What about $\pi$ and $e$ and $\log 2$?
Are all of these sequences "random" in the same sense as the Liouville and Moebius functions?
David Speyer is right. A Dirichlet series with coefficients that are random choices from $0,1$ does not count as a random Dirichlet series. Random choices from $-1,1$ would count as a random Dirichlet series. Taking his insight into account, I correct my prediction: With probability $1$ the Dirichlet series $\sum b_n(x)n^{-s}$ where $b_n(x)$ are the binary digits of $x \in [0,1)$ will have a meromorphic continuation to $\sigma > 1/2$ with a single simple pole at $s = 1$ with residue $1/2$. Also I now remember a reference: Armand Denjoy L'Hypothese de Riemann sur la distribution... C. R. Acad. Sci. Paris 192 (1931), 656-658. This paper deals with the situation where the Dirichlet series has coefficients chosen at random from $-1,1$.
A random Dirichlet series has its line of convergence as a natural boundary with probability $1$. The particular series coming from $\sqrt{2}$ that you are looking at is overwhelmingly unlikely to have an analytic continuation beyond $\sigma > 1$ (Correction: I was too hasty; the series is likely to have a meromorphic continuation to $\sigma > 1/2$. See the objection by David Speyer and my corrected answer.) The Law of Large Numbers does not show that almost all reals in $[0,1)$ correspond to Dirichlet series that converge for $\sigma > 1/2$. In fact, the set of reals for which the summatory function of the binary digits is $o(x)$ obviously is a null set. So the line of convergence is $\sigma = 1$ with probability $1$ and this is then a natural boundary with probability $1$. (Correction: In this situation with random choices from $0,1$, $\sigma = 1$ will be the line of convergence, but not a natural boundary, with probability $1$. See the objection by David Speyer and my corrected answer.)
You have been misled by the analogy with the Moebius function. The latter oscillates around zero, which your binary expansion coefficients do not.
By the way, the Law of Large Numbers is too weak to draw the kind of conclusions that you want. You need something stronger, for example the Law of the Iterated Logarithm.
I would expect that the Dirichlet series with coefficients $c_n = 2b_n - 1$ where $b_n$ is the n-th binary digit of $\sqrt{2}$ should converge in $\sigma > 1/2$ and have $\sigma = 1/2$ as a natural boundary.
It is not known whether pi has a random distribution of digits in its expansion. Here is an article on this and related problems:
