Simply normal sequence of $-1$'s and $1$'s as coefficients of harmonic series Suppose $s_{n}$ is either $1$ or $-1$ for $n=1,2, 3,\ldots$ and that half the $s_{n}$'s are 1; i.e. 
$$
\lim_{n\to \infty} \frac{\#\{i\leq n: s_{i}=1\}}{n}=\frac{1}{2}.
$$
Does then the series $\sum_{i=1}^{\infty} \frac{s_{n}}{n}$ converge?

The point here is that the alternating harmonic series converges, and if the $s_n$'s are the result of a fair coin flip ($-1$ for heads and $1$ for tails), then the series converges almost surely. So I'm asking about convergence for a simply normal sequence of $1$'s and $-1$'s. The sequence $1,-1,1,-1,1,-1,\ldots$ is simply normal, as is the result of the coin flip (almost surely).
I just found this paper in which there's an exercise at the end claiming that "Balancing the plus and
minus signs does not guarantee convergence."
Well I pretty sure that your series converges :)

So... as far as i understand, $s_n$ is the sequence which for each $n$ gives you either $1$ or $-1$, but not neccesarily in the order $1,-1,1,-1,1...$ Is that right? 

If the sequence is $s_n=(-1)^n$ then of course the series $\sum_{k=1}^n\frac{s_n}{n}$ converges. Here is a link to the alterating series test: http://www.math.com/tables/expansion/tests.htm
whether it converges to $\frac{1}{2}$ i can't say for sure but you could try to make the deference between $\frac{1}{2}$ and your series up til $n$ less then an arbitrary number, $\epsilon$. In that way you would have proved that the series equals $\frac{1}{2}$ 

But I'm not completely sure if we're just looking at random series with mixed up $1$ and $-1$...