Doubt in exercise 3.1.12 of Book Problems in analytic number theory by My Ram Murthy

Question

While trying problems from above mentioned book I am unable to think about how to prove the question which I am writing below.

Question is ->If limit x->$\infty \frac {π(x) } {x/log(x) } $ = $\alpha$ Then show that $\sum_{p\leq x} 1/p = \alpha log log(x) + o(log log(x) ) $ .

What I thought ->taking a(n) = 1 if n = prime, 0 otherwise and f(n) = 1/n and using abel summation formula I got $\sum_{p\leq x } 1/p = π(x) / x + \int_{2}^x \frac {π(x) } {t^2} dt $ .

Now using $\int_{2}^x = \int_{2}^{\infty} -\int_x^{\infty} $ I get π(x) = O(1/log (x) ) + O $(\int_2^{\infty} \frac {1} {t logt } dt - \alpha \int_x^ {\infty} \frac {1} { t logt } dt )$ .

Now the problem is $ loglog (\infty) $ diverges.

Can someone please tell where I am doing mistake in integral.

I tried this problem yesterday also but couldn't solve it. Please help.

Answer 1

You are applying the partial summation in a completely wrong way. $$\sum_{p\le x} 1/p=\sum_{n\le x} \frac{\pi(n)-\pi(n-1)}{n}=\frac{\pi(x)}{x}+\sum_{n\le x-1} \pi(n)(\frac1n-\frac1{n+1})$$ $$= \frac{\frac{x}{\log x}a(1 +o(1))}{x}+\sum_{n\le x-1}\frac{n a(1 +o(1))}{\log n} \frac{1+o(1)}{n^2}= a(1+o(1))\log \log x$$ (the last step is because the derivative of $\log \log x$ is $\frac1{x\log x}$)

Note that since $\ell=\sum_p 1/p^2$ converges, we do it the opposite way : $$\sum_{p< x} 1/p^2=\ell-\sum_{p\ge x}1/p^2=\ell-\frac{\pi(x)}{x^2}-\sum_{n\ge x} \pi(n)(\frac1{n^2}-\frac1{(n+1)^2})$$ The integral version is the same, just replace $\frac1n-\frac1{n+1}=\int_n^{n+1} \frac1{t^2}dt$

Answer 2

This will be the same solution in a different language... I like using Stieltjes integrals and integrating by parts better than Abel transforms (which is the same): $$ \sum_{p\le x} \frac1p = \int_{2-}^x \frac{\mathrm{d}\pi(t)}{t} = \left[\frac{\mathrm{d}\pi(t)}{t}\right]_{2-}^x-\int_{2-}^x \pi(t)\mathrm{d}\left(\frac1t\right) = \frac{\pi(x)}{x}+\int_{2}^x\frac{\pi(t)}{t^2} \mathrm{d}t. $$ After this point we just have to replace $\pi(t)$ by $\alpha\frac{t}{\log t}$, knowing that $\int_2^x\frac{\mathrm{d}t}{t\log t} = \log\log x+O(1)$: $$ \left|\sum_{p\le x} \frac1p-\alpha\log\log x\right| = \left|\int_{2}^x\frac{\pi(t)}{t^2} \mathrm{d}t -\int_{2}^x\frac{\alpha\frac{t}{\log t}}{t^2} \mathrm{d}t +O(1)\right| \le \int_{2}^x\frac{\Big|\frac{\log t}{t}\pi(t)-\alpha\Big|}{t\log t} \mathrm{d}t+O(1). $$ For every fixed $\varepsilon>0$ there is some $K=K(\varepsilon)$ such that $\Big|\frac{\log t}{t}\pi(t)-\alpha\Big|<\varepsilon$ for $t\ge K$, so $$ \int_{2}^x\frac{\Big|\frac{\log t}{t}\pi(t)-\alpha\Big|}{t\log t} \mathrm{d}t =\int_2^K+\int_K^x \le O_\varepsilon(1)+\int_K^x\frac{\varepsilon}{t\log t} \mathrm{d}t = \varepsilon\log\log x + O_\varepsilon(1); $$ $$ 0 \le \limsup_{x\to\infty} \left|\frac{\sum_{p\le x} \frac1p}{\log\log x}-\alpha\right| \le \limsup_{x\to\infty} \left(\varepsilon+\frac{O_\varepsilon(1)}{\log\log x}\right) =\varepsilon $$ This holds for all $\varepsilon>0$, so $$ \lim_{x\to\infty} \left|\frac{\sum_{p\le x} \frac1p}{\log\log x}-\alpha\right| = 0. $$