The sum ∞∑n=21nln(n) does not converge.
But the sum ∞∑n=11Pnln(Pn) where Pn denotes the nth prime number appears to be.
Is that correct, and if so, how can we calculate the value of convergence?
Is it possible that this sum converges to the golden ratio (1+√52)?
Answer
With Pn≈nln(n), we should have ∞∑N1Pnln(Pn)≈∫∞Ndxxln(x)2=1lnN
If the sum for n up to π(19999999)=1270607 is 1.57713, we'd expect
the remainder to be about .071, which would push the total to about 1.648, too high for ϕ.
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