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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 $\pi$.

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    Is there a proof that the sum converges ? – 2017-01-28
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    If the sum diverges to infinty, are you still writing that it is $>0$? – 2017-01-28
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    Andre S. I suspect no and I never thought of that. Is there a reason to suspect that at arbitrarily large values of $n$ the finite sum might be negative ? – 2017-01-28
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    For all we know, all digits of $\pi$ after position $10^{20}$ may be $2$, $3$ or $4$ ... – 2017-01-28
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    The series diverges since there is some $m\in \{0,1,\dots ,9\}$ such that $\pi_n=m$ for infinitely many $n,$ proving $\cos (\pi_n)$ does not converge to $0.$ – 2017-01-28

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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.

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    The actual sum I have is $20,680.311216004\ldots $ – 2017-01-28
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    Using your reasoning at $1,000$ we expect $\gamma=42$ but it is closer to $29$. At $2,000$ we expect $\gamma=84$ but it's actual sum is closer to $71$. – 2017-01-28
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    @AnthonyHernandez - If $\pi$ is normal, it doesn't say that the digits are uniformly distributed on any interval, just that on average they are. The more digits you sum, the closer you'd expect the numbers to match up. – 2017-01-28
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    Selecting a random number uniformly from $\{\cos 0,\cos 1,\dots,\cos 9\}$ has a variance of roughly [0.52768](http://www.wolframalpha.com/input/?i=variance+of+%7Bcos+0,+cos+1,+cos+2,+cos+3,+cos+4,+cos+5,+cos+6,+cos+7,+cos+8,+cos+9%7D&rawformassumption=%7B%22C%22,+%22variance%22%7D+-%3E+%7B%22PopulationVariance%22%7D&rawformassumption=%22TrigRD%22+-%3E+%22R%22). So, if $\pi$ is normal, the error in this estimate after $d$ digits should be on the order of $\sqrt{0.52768d}$. When $d=1000$, this is roughly 23; when $d=2000$, it's roughly 32; when $d=480000$, it's roughly 503. – 2017-01-28
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    In all three cases, the error in the estimate is of the size you'd expect (assuming $\pi$ is normal). – 2017-01-28