An intuitive reason why the $\sum_{n=1}^\infty \cos(\pi_n)$ might be positive where $\pi_n$ is the n-th digit of $\pi$

Question

Let $\pi_n$ be the $n-th$ digit in the base ten decimal representation of $\pi$. Is there an intuitive reason why it might be true $$\sum_{n=1}^\infty \cos(\pi_n)>0$$ For short handedness I write $\gamma = \sum_{n=1}^\infty \cos(\pi_n)$. Over the first 480,000 digits of $\pi$ I have that $\gamma \approx 20,000$. Surely $0 \leq \pi_n \leq 9$ and $\cos(\pi_n)<0$ if $\pi_n$ is one of $2,3,4,8\text{ or }9.$ The big jumps in the sum occur whenever $\pi_n=0$ in which case $\cos(\pi_n)=1$. It's a reach for me but I cannot think of a reason why this sum might be positive or negative. I am not able to compute this sum for the first 1,000,000 digits of $\pi$ so I am not certain that $\gamma$ "turns" back to negative. Moreover I am not sure if it is true $\gamma \neq 0$. This is just pure curiosity and fun with the digits of $\pi$.

Answer 1

Averaging over the digits $\{0\ldots9\}$, the value $\cos(\pi_n)$ is on average larger than zero, at $\approx0.042$. If $\pi$ is even close to being a normal number (and there is currently no evidence to suppose otherwise), you would expect that the sum over the first $480,000$ digits would be around $480,000\times0.042 \approx20160$, which is pretty close to what you got.

Of course, it's still entirely possible that $\pi$ becomes a long stream of $3$'s at some point, and your sum will be negative - but to the best of my knowledge, we just don't know.