Asymptotics of the n-th prime using the gamma function

In the paper http://rgmia.org/papers/v8n2/eepnt.pdf, the author proves that proves an explicit inequality on prime numbers using the gamma function and as a corollary, he showed that.

$$p_n = n \frac{\Gamma'(n)}{\Gamma(n)} + o(n \ln n).$$

I obtained a stronger form of this result namely

$$p_n = n \ln \frac{\Gamma'(n)}{\Gamma(n-1)} + O\Big(\frac{n\ln\ln n}{\ln n}\Big).$$

The gamma function seems to beautifully approximate $p_n$. To get the same error term using the regular Cipolla's asymptotic expansion of the $p_n$ we would need three terms.

Can someone explain why the gamma function approximated the n-th prime so nicely? Is this a coincidence or is there some underlying phenomenon governing this result that can shed some new light distribution of prime numbers.

• Something is wrong here. $\Gamma'(n)/\Gamma(n) \approx \log n$ so you formula gives $p_n \approx n \log \log n$, where the truth is $p_n \approx n \log n$. – David E Speyer Jun 1 '12 at 12:02
• @ David. It was a typing error and I have corrected it. – Nilos Jun 1 '12 at 13:03

The asymptotic expansion of Cipolla starts $$p_n=n\log n+n\log\log n-n+n\frac{\log\log n}{\log n}+O(n(\log\log n/\log n)^2)$$ So the given approximations have errors $$p_n=n\frac{\Gamma'(n)}{\Gamma(n)}+\Theta(n\log\log n)$$ and $$p_n=n\log\frac{\Gamma'(n)}{\Gamma(n-1)}+\Theta(n).$$ I would not say these are good approximations with so big errors.

The inverse function of the log integral function $\text{li}^{-1}(x)$ has error $$p_n= \text{li}^{-1}(n) +O(n \exp(-c\sqrt{\log n})$$ which assumming Riemann hypothesis can be reduced to $$|p_n-\text{li}^{-1}(n)|\le \pi^{-1} \sqrt{n}(\log n)^{\frac52}\qquad n>11.$$ (see arXiv:1203.5413)

Can someone explain why the gamma function approximated the n-th prime so nicely?
Is this a coincidence or is there some underlying phenomenon governing this result?

The expression which appears there is $\dfrac{\Gamma'(n)}{\Gamma(n)}=\Big[\ln\Gamma(n)\Big]'=\psi_0(n)$, see digamma function.

This famous function has the property that $\psi_0(n)=H_{n-1}-\gamma$, where $H_n$ is the n-th harmonic

number
, and $\gamma\simeq\dfrac1{\sqrt3}$ is the Euler-Mascheroni constant. But, at the same time, $H_{n-1}\simeq\ln n$,

the small error or difference between the two tending towards the afore-mentioned constant. By

differentiating the natural logarithm of the fundamental property of the $\Gamma$ function, $\dfrac{\Gamma(n+1)}{\Gamma(n)}=n$,

we have that $\psi_0(n+1)=\dfrac1n+\psi_0(n)$, which is nearly identical to the recurrence relation of

harmonic numbers, $H_{n+1}=\dfrac1{n+1}+H_n$. The only difference is the offset, $\psi_0(1)=-\gamma\neq$

$\neq H_0=0$, a proof of which can be found here. The conclusion then follows from the prime number theorem.