TITLE: Asymptotics of the n-th prime using the gamma function
QUESTION [13 upvotes]: 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.

REPLY [7 votes]: 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.