# Partial sum square root of reciprocal of primes

I would like to know if the following reasoning makes sense. I want to bound/estimate the following sum $$\sum_{p\leq x}\frac{1}{\sqrt{p}}$$ Using integration by parts we have \begin{align} \sum_{p\leq x}\frac{1}{\sqrt{p}}&=\int_2^x \frac{1}{\sqrt{t}}\,d(\pi(t))\\ &=\left[\frac{\pi(t)}{\sqrt{t}}\right]_2^x+\frac{1}{2}\int_2^x\frac{\pi(t)}{t^{3/2}}\,dt\\ &=\frac{\pi(x)}{\sqrt{x}}+\frac{1}{2}\int_2^x\frac{\pi(t)}{t^{3/2}}\,dt \end{align} Now, using that by the PNT we have $$\pi(x)\sim \dfrac{x}{\ln x}$$, we get $$\sum_{p\leq x}\frac{1}{\sqrt{x}}\sim\frac{\sqrt{x}}{\ln x}+\frac{1}{2}\int_2^x\frac{1}{\sqrt{t}\ln t}\,dt$$ On the other hand we have $$\int_2^x \frac{1}{\sqrt{t}\ln t}\,dt=\operatorname{Ei}\left(\frac{\ln x}{2}\right)-\operatorname{Ei}\left(\frac{\ln 2}{2}\right)\sim\operatorname{Ei}\left(\frac{\ln x}{2}\right)\sim\frac{2\sqrt{x}}{\ln x}$$ which I got using wolframalpha. Hence I obtain $$\sum_{p\leq x}\frac{1}{\sqrt{p}}\sim \frac{2\sqrt{x}}{\ln x}$$ Does it make sense? How could I prove the asymptotic for $$\int_2^x\frac{1}{\sqrt{t}\ln t}\,dt$$ without the need to refer to wolframalpha? Thank you!

• One direction: you can lower bound the integral by replacing log t by log x and you get 2sqrt(x)/log(x). Now you can similarly fiddle around to get a good upper bound – Sandeep Silwal Jul 9 at 1:39

If $$\pi(x)$$ is the number of primes not greater than $$x$$, then $$\pi(x)$$ is continuous from the right and the Riemann-Stieltjes integral over $$[2, 2 + \epsilon]$$ will tend to zero. The first equation should be $$\sum_{p \leq x} \frac 1 {\sqrt p} = \frac 1 {\sqrt 2} + \int_2^x \frac {d \pi(t)} {\sqrt t} = \frac {\pi(x)} {\sqrt x} + \frac 1 2 \int_2^x \frac {\pi(t)} {t^{3/2}} dt.$$
To prove the asymptotic equivalence of the integrals, show that l'Hopital's rule applies. Then $$\lim_{x \to \infty} \frac {\int_2^x t^{-3/2} \, \pi(t) \, dt} {\int_2^x t^{-1/2} \ln^{-1} t \, dt} = \lim_{x \to \infty} \frac {x^{-3/2} \, \pi(x)} {x^{-1/2} \ln^{-1} x} = 1.$$ To estimate the integral in the denominator, apply integration by parts and l'Hopital's rule again: $$\frac 1 2 \int_2^x \frac {dt} {\sqrt t \ln t} = \frac {\sqrt t} {\ln t} \bigg\rvert_{t = 2}^x + \int_2^x \frac {dt} {\sqrt t \ln^2 t} = \frac {\sqrt x} {\ln x} + o {\left( \frac {\sqrt x} {\ln x} \right)}.$$ Therefore your final result is correct.