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

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.

## 2 Answers

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 $$\frac1n-\frac1{n+1}=\int_n^{n+1} \frac1{t^2}dt$$

• M ram Murthy solution is same as mine up to the point of calculating integrals. Also, I am following approach of Apostol – Ben Jan 16 '20 at 7:54
• can you please help me with this problem if you have some spare time. My presentation of it is due Saturday!! math.stackexchange.com/questions/3543021/… – Ben Feb 12 '20 at 18:47
• Hi, I told you,the partial summation of series, finite sum and Stieltjes integrals is the same thing: $b(n)-b(n+1)=\int_n^{n+1} b'(t)dt$ so $\sum_n (\sum_{m\le n} a(m)) (b(n)-b(n+1)) = \sum_n (\sum_{m\le n} a(m))\int_n^{n+1} b'(t)dt=\int_1^\infty (\sum_{m\le t} a(m)) b'(t)dt$. The main problem in your question was that you didn't notice that when $\sum_n a(n) b(n)$ converges we apply the partial summation to $\sum_{n\ge N}a(n)b(n)$, not to $\sum_{n\le N} a(n)b(n)$, when it diverges we apply the partial summation to $\sum_{n\le N} a(n)b(n)$, in order to find its rate of convergence/divergence. – reuns Feb 12 '20 at 18:54
• With $A(t)=\sum_{n\le t} a(n)$ then $\sum_{n\le N} a(n)b(n) = \int_{1-\epsilon}^{N+\epsilon} b(t) d(A(t))=b(t) A(t)|_{1-\epsilon}^{N+\epsilon}-\int_{1-\epsilon}^{N+\epsilon} A(t) d(b(t))$. – reuns Feb 12 '20 at 18:58
• I am really sorry the question I wanted to ask is not the above asked question to which you answered ie the problem in Ram Murthy book but the question whose link I mentioned in comments. I am really sorry that you thought there is some doubt again in this question. I have understood it. I am really sorry for wasting your precious time but the question for which I wanted your help is - > math.stackexchange.com/questions/3543021/… . If you have some spare time can you please help. I am really sorry. – Ben Feb 12 '20 at 19:21

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 $$\lim_{x\to\infty} \left|\frac{\sum_{p\le x} \frac1p}{\log\log x}-\alpha\right| = 0.$$

• thank you very much for answering. I couldn't see the answer. I am very busy and bounty expires in 2 hours. If I have any question in understand the answer, I will mention to you in comments. Please try to understand.i am awarding you the bounty. – Ben Jan 30 '20 at 12:43