Asymptotics of sum of Chebychev function

Question

Show that $\sum_{n\leq x}\frac{θ(n)}{n^2}=\ln x+O(1)$ where $θ$ is the Chebychev function. (We are searching for a solution without the prime number theorem, just Chebychev bounds or something like that, but the exercise is talking about a solution without the PNT.)

First my thought was to maybe use Shapiro for the fucntion $θ(n)/n$, but it doesn't seem to work.

Then I thought to use Shapiro together with the fact that $\sum_{n\leq x}θ(x/n)=x\log x+O(x)$ or the fact that $\sum_{n\leq x}ψ(x/n)=x\log x+O(x)$ so as to bring $\sum_{n\leq x}\frac{θ(n)[x/n]}{n}$ close to one of the 2 last sums (like minus $O(x)$ or $O(1)$) and then use Shapiro, but I cant find something clear. Any ideas, suggestions?

Answer 1

• $\sum_{p\in (n,2n]} \log p \le \log {2n\choose n}=O(n)$ gives that $\sum_{p^k\le n} \log p =O(n)$

• So $n \log n +O(n)= \log n! = \sum_{p^k\le n} \lfloor n/p^k \rfloor \log p= \sum_{p^k\le n} \frac{n }{ p^k}\log p +O(n)$ $= \sum_{p\le n} \frac{n }{ p}\log p+O(n)$

• With a partial summation we get $O(1)+\log x=\sum_{p\le x}\frac{\log p}p= \frac{\theta(n)}n+\sum_{n\le x-1}\theta(n)(\frac1n-\frac1{n+1})= O(1)+\sum_{n\leq x}\frac{\theta(n)}{n^2}$

Answer 2

This is just an expanded account of the method suggested by Greg Martin:

\begin{aligned} \sum_{n\le x}{\vartheta(n)\over n^2} &=\sum_{n\le x}{1\over n^2}\sum_{p\le n}\log p=\sum_{p\le x}\log p\sum_{p\le n\le x}{1\over n^2} \\ &=\sum_{p\le x}\log p\left\{\int_p^x{\mathrm dt\over t^2}+O\left(1\over p^2\right)\right\} \\ &=\sum_{p\le x}{\log p\over p}+O\left\{\sum_{p\le x}{\log p\over p^2}\right\}+O\left\{\frac1x\sum_{p\le x}\log p\right\} \\ &=\sum_{p\le x}{\log p\over p}+O(1)=\log x+O(1), \end{aligned}

wherein the second last inequality follows from Chebyshev's bound, and the last equality follows from Mertens' first theorem.