I came across a good question - maybe it was answered before, if so please point me to it.
Consider the series 1/a_n where a_n = n th digit of pi. So, S(n), the sum of the series :
S(n)=1/3 + 1/1 + 1/4 +1/1+1/5…+1/a_n
Is there an analytic representation of the sum to n terms ? I am guessing the series does not converge - is there a proof ?
Thanks
(Assuming you remove all terms which would result in 1/0…)
Well, it trivially won’t converge if pi is normal, because then there will be an infinite number of 1/1 terms, so it will keep trending upwards indefinitely.
However, we don’t know if pi is normal.
On the gripping hand, pi being normal is a sufficient but not necessary condition for non-convergence: It is very easy to construct non-normally-distributed infinite sequences which won’t converge if added in the way you propose.
So I don’t see any special reason your sum would converge, and since there’s so many ways for it to diverge, I’m pretty sure it’s going to be divergent.
I can think of a way to formulate the summation, but it involves modular arithmetic.
sum over i=0->n { 1/(floor(π*10[sup]i[/sup]) mod 10) }
And assuming it’s passed through a function that removes 0 values.
I don’t have enough of a background in integer series to be able to analyze this, but I suspect it’s either divergent or (currently) unprovable.
There’s probably a better way to formulate this question in terms of one of the infinite series that evaluate to pi.
No. A series 1/9 + 1/9 + 1/9 + … will diverge if it has an infinite number of non-zero terms, so yours will also. And the decimal expansion of pi contains an infinite number of non-zeros; otherwise it would eventually be …0000… forever which happens only for rational numbers.