Showing that $p_n = n\log n + n\log \log n + O(n)$ (simplified prime approximation) This StackExchange question -- Is the $n$-th prime smaller than $n(\log n + \log\log n-1+\frac{\log\log n}{\log n})$? -- assumes the following statement: 
$$\log n + \log\log n -1 \leq \frac{p_n}{n} \leq \log n + \log\log n.$$ 
My question is: how can we prove that statement? 
Using the version of the Prime Number Theorem that states $$\pi (n) \sim \frac{n}{\log n},$$ we can quite easily show that $$p_n \sim n\log n,$$ or in other words $$p_n = n\log n + o(n\log n),$$ but this doesn't seem to get us any closer to a big-O estimate of the kind assumed in the above question. Is there a standard, quick method of obtaining such an estimate?
 A: The question you are asking especially the lower limit was proved by Pierre Dusart using the first 3.5 million zeroes of the Riemann zeta function so don't expect an easy answer. Here is the full solution. 
The $k$-th prime is greater than $k(\ln k + \ln \ln k − 1)$ for $k \ge 2$
A: (Answering my own question because I was given the solution elsewhere.) 
By a known result and simple logarithm laws we have
$$\begin{align} p_n &\sim n\log n \\ \log p_n &= \log n + \log \log n + o(1).   \end{align}$$ 
We also know that $\pi (n) = \frac{n}{\log n} + O\left(\frac{n}{(\log n)^2}\right),$ and therefore $$n = \pi (p_n) = \frac{p_n}{\log p_n} + O\left(\frac{p_n}{(\log p_n)^2} \right).$$ 
Meanwhile $O\left(\frac{p_n}{(\log p_n)^2} \right)$ simplifies, using the little-o estimates above, to $O\left(\frac{n}{\log n} \right)$. Plugging this in gives $$n= \frac{p_n}{\log n +\log\log n + o(n)} +O\left(\frac{n}{\log n} \right),$$ 
and therefore, just by rearranging terms and cancellation, $$p_n = n\log n + n\log\log n + O(n).$$
