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).
 A: 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."
A: As the paper you found indicates, not every simply normal sequence of $-1$'s and $1$'s makes the series converge. For example, define the following sequence:
$$
s_n=\left\{\begin{array}{ll}
1 & : n \textrm{ is prime} \\
(-1)^n & : \textrm{otherwise}
\end{array}\right.
$$
That is, we alternate between $-1$ and $1$, except for the prime terms, which are always $1$. Then, for simple normality, we have $\displaystyle\lim_{n\rightarrow\infty}\frac{N_s(1,n)}{n}=\lim_{n\rightarrow\infty}\frac{\frac{n}{2}+\pi(n)}{n}=\frac{1}{2}$, where $\pi$ is the prime counting function. Since $\displaystyle\sum_{p\textrm{ is prime}}\frac{1}{p}$ diverges, we know that $\displaystyle\sum_{n=1}^{\infty} \frac{s_n}{n}$ diverges as well.
So we can't say with certainty that the series converges.
However, since almost all real numbers are (absolutely) normal, it follows that almost all simply normal numbers are absolutely normal. Since it is easy to construct a bijection between $[0,1]$ and all seqences of $-1$'s and $1$'s (take the binary decimal, each digit is $0$ or $1$, corresponding to $-1$ and $1$ in the series, respectively), almost all of your sequences $\{s_n\}$ are absolutely normal. As the alternating sequence shows, a simply normal sequence that is not absolutely normal can still lead to a converging series. (I would guess that almost all of them do, but I'm not sure). Since an absolutely normal sequence corresponds to a random harmonic series, the series converges almost surely.
A: 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 $-1$... 
