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I have observed, by calculation, that the sum of the reciprocals of all the n-digit prime numbers is approximately 1/n, and that this becomes increasingly accurate as n increases. Is there a simple way to prove this?

I have also observed that this appears to hold true whether one is working in base 10, or in base 2, or in any other base. Again, is there a simple way to prove this?

John Grint
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Sum of reciprocals of prime numbers less than $m$ is roughly $f(m ) = \ln( \ln(m))$ up to a constant (I don't know any elementary proof of this fact though) .Now note that $n$-digit numbers are just numbers between $10^{n-1}$ and $10^n$, so you are looking for $f(10^n) - f(10^{n-1})$. This doesn't quite match with your observation though.

Mehdi
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  • It kind of matches. The $\ln\ln$ of $10^n$ is $\ln k+\ln n$ where $k$ is the conversion factor. So your difference is $\ln(n/(n-1))$ which is about $1/(n-1)$. – André Nicolas Feb 04 '16 at 04:15
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    The proof comes from the prime number theorem. A number $k$ has $\frac 1{\log k}$ chance of being prime, so instead of adding the prime reciprocals add up the reciprocals of all the numbers times $\frac 1{\log k}$. If you convert that to an integral, you get $\int_{10^{n-1}}^{10^n}\frac dx{x \log x}$. The integral is just $\log (\log x)$ – Ross Millikan Feb 04 '16 at 04:58
  • @AndréNicolas Good point , the difference is of order 1/n^2 which is small comparing to the value 1/(n-1) – Mehdi Feb 04 '16 at 20:04