# Does $\sum\limits_{n=1}^{\infty}\frac{1}{P_n\ln(P_n)}$ converge to the golden ratio?

The sum $\displaystyle\sum\limits_{n=2}^{\infty}\frac{1}{n\ln(n)}$ does not converge.

But the sum $\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{P_n\ln(P_n)}$ where $P_n$ denotes the $n$th 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 ($\dfrac{1+\sqrt{5}}{2}$)?

• That series does in fact converge since $p_n\sim n\log n +o(n\log n)$ (i.e. the Prime Number Theorem). Why would you think it converges to $\phi$? – Adam Hughes Aug 5 '14 at 22:13
• If this is true I would be intensely surprised. – Adam Hughes Aug 5 '14 at 22:14
• I'd be shocked if this converged to anything that has a name, let alone a quadratic irrational like $\phi$. – Semiclassical Aug 5 '14 at 22:29
• You must have an error, I just computed the sum, its value is Pi/2. Terence Tao would approve. – TROLLHUNTER Aug 5 '14 at 22:53
• @SandeepSilwal I think it's supposed to be a joke. – Slade Aug 6 '14 at 0:28

With $P_n \approx n \ln(n)$, we should have $$\sum_{N}^\infty \dfrac{1}{P_n \ln(P_n)} \approx \int_N^\infty \dfrac{dx}{x \ln(x)^2} = \dfrac{1}{\ln N}$$ If the sum for $n$ up to $\pi(19999999) = 1270607$ is $1.57713$, we'd expect the remainder to be about $.071$, which would push the total to about $1.648$, too high for $\phi$.
• Can you please explain the (capital) $N$ in your formula? My sum starts at $1$, so if that's what you meant then $\frac{1}{\ln1}$ is quite undefined. – barak manos Aug 6 '14 at 4:15
• @barak manos, for large $N$ one can approximate $$\sum_{n=1}^\infty \dfrac{1}{P_n \ln(P_n)} \approx \sum_{n=1}^{N-1} \dfrac{1}{P_n \ln(P_n)} + \sum_{n=N}^\infty \dfrac{1}{n \ln(n) \cdot \ln\left( n \ln(n)\right)} \approx \sum_{n=1}^{N-1} \dfrac{1}{P_n \ln(P_n)} + \dfrac{1}{\ln N},$$ which is $\pm$ appropriate estimation. – Oleg567 Aug 6 '14 at 5:48
• @Oleg567: So the sum diverges????? (since $N=\infty$). – barak manos Aug 6 '14 at 5:51
• @barakmanos, why diverges? To estimate series more accurate, we can choose $N$ bigger and bigger. Series is convergent definitely. – Oleg567 Aug 6 '14 at 5:53
• @barakmanos, 1st step: to take $N=10$ (for instance), 2nd step: to take $N=100$, and so on ... $\dfrac{1}{\ln N} \rightarrow 0$, when $N\rightarrow +\infty$. – Oleg567 Aug 6 '14 at 5:55