Approximating $\vartheta(x)=\sum_{p\le x} \log(p)?$

Question

Consider the first Chebyshev function $\vartheta(x)=\sum_{p\le x} \log(p)$ where the sum runs over the primes less than or equal to $x$.

I wanted to approximate $\vartheta(x).$

My attempt was $f(x)=\sum_{n \ge 2}^x n^\frac{-1}{n}.$ It overcounts by about $2$ at $x=101$ giving a value of $90.177$ whereas $\vartheta(x)$ gives $88.344.$ I'm not sure how $f(x)$ performs as $x$ increases.

Is $f(x)\sim \vartheta(x)?$

Answer 1

Per $$n^{-1/n} = \exp \biggl(-\frac{\log n}{n}\biggr) = 1 - \frac{\log n}{n} + O\biggl(\frac{(\log n)^2}{n^2}\biggr)$$ we have $$f(x) = x - \frac{1}{2}(\log x)^2 + O(1)\,.$$ Thus we have $f(x) \sim \vartheta(x) \sim x$.

But $f(x)$ stays much closer to $x$ than $\vartheta(x)$. By a result of Littlewood we have $$\vartheta(x) - x \in \Omega_{\pm}(\sqrt{x}\, \log \log \log x)$$ while $x - f(x)$ is of much smaller magnitude, and always positive.