14
$\begingroup$

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 ($\dfrac{1+\sqrt{5}}{2}$)?

$\endgroup$
  • $\begingroup$ 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 $\phi$? $\endgroup$ – Adam Hughes Aug 5 '14 at 22:13
  • 7
    $\begingroup$ If this is true I would be intensely surprised. $\endgroup$ – Adam Hughes Aug 5 '14 at 22:14
  • 7
    $\begingroup$ I'd be shocked if this converged to anything that has a name, let alone a quadratic irrational like $\phi$. $\endgroup$ – Semiclassical Aug 5 '14 at 22:29
  • 3
    $\begingroup$ You must have an error, I just computed the sum, its value is Pi/2. Terence Tao would approve. $\endgroup$ – TROLLHUNTER Aug 5 '14 at 22:53
  • 3
    $\begingroup$ @SandeepSilwal I think it's supposed to be a joke. $\endgroup$ – Slade Aug 6 '14 at 0:28
12
$\begingroup$

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 $\phi$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – barak manos Aug 6 '14 at 4:15
  • 1
    $\begingroup$ @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. $\endgroup$ – Oleg567 Aug 6 '14 at 5:48
  • $\begingroup$ @Oleg567: So the sum diverges????? (since $N=\infty$). $\endgroup$ – barak manos Aug 6 '14 at 5:51
  • $\begingroup$ @barakmanos, why diverges? To estimate series more accurate, we can choose $N$ bigger and bigger. Series is convergent definitely. $\endgroup$ – Oleg567 Aug 6 '14 at 5:53
  • $\begingroup$ @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 $N\rightarrow +\infty$. $\endgroup$ – Oleg567 Aug 6 '14 at 5:55
5
$\begingroup$

1.63661632335... See http://oeis.org/A137245 and links therein (also http://en.wikipedia.org/wiki/Prime_zeta_function, scroll down to integral section)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.