It has been written (in the article "A discussion of the fundamental ideas behind Selberg’s Elementary proof of the prime-number theorem” by Steve Balady, page 7) that,

enter image description here

On page 10, it is written - enter image description here

But I can not see how $(6.2)$ is obtained using $(4.5)$ from the previous line $$\sum_{d \leq x}\mu(d) \sum_{d' \leq \frac x d }\log^2(d')$$,

I tried by simplifying (brute force, multiplication) but didn't yield the result,

Can any one kindly show that plz?


1 Answer 1


First, there is a typo in the statement (6.2): the first sum should be multiplied by $x$. We can see that this is merely a typo by looking at how they use (4.10) right under.

With that in mind, it suffices to prove that $$ \frac12 \sum_{n\le x}\log^2 n = \frac12 x\log^2 x - x\log x + x+ O(\log^2 x) $$ and apply it to the sum $\sum_{d'\le\frac xd}\log^2 d'$.

I assume for convenience that $x$ is an integer. Look at (4.5) and take a sum over the whole equation: $$ \frac12 \sum_{n\le x}\log^2 n = \sum_{n\le x}\sum_{m\le n}\frac{\log m}{m} - xC'+ O\left(\sum_{n\le x}\frac{\log n}{n}\right) \tag{1} \label{1} $$ The last term is $O(\log^2x)$ by (4.5). The first term is equal to $$ \begin{split} \sum_{m=1}^x\sum_{n=m}^x\frac{\log m}{m} &= \sum_{m=1}^x(x-m+1)\frac{\log m}{m}\\ &= (x+1)\sum_{m=1}^x\frac{\log m}{m} - \sum_{m=1}^x\log m\\ &= \frac12(x+1)\log^2 x + (x+1)C' + O\left(\log x\right) - \sum_{m=1}^x\log m \end{split} \tag{2} \label{2} $$ Here we used (4.5) again. I claim that $$ \sum_{m=1}^x\log m = x\log x - x + O(\log x) \tag{3} \label{3} $$ Plug \eqref{2} and \eqref{3} into \eqref{1}, and you get the result.

To prove \eqref{3}, you can use the same strategy on (4.4) instead of (4.5) in the article. The proof is completely analogous.

Let me know if you need more details on something :)


Explanation of \eqref{2}, line 1: First we swap the order of summation, $\sum_{n=1}^x\sum_{m=1}^n = \sum_{m=1}^x\sum_{n=m}^x$, giving the left hand side of \eqref{2}. The terms in the sum are constant with respect to $n$, and $x-m+1$ is the number of terms in the inner sum, so the inner sum works out to be $$ \sum_{n=m}^x\frac{\log m}{m} = \frac{\log m}{m}\sum_{n=m}^x 1 = \frac{\log m}{m} (x-m+1) $$

  • $\begingroup$ The following line is not clear , plz explain elaborately:$ \sum_{m=1}^x\sum_{n=m}^x\frac{\log m}{m} = \sum_{m=1}^x(x-m+1)\frac{\log m}{m} $ $\endgroup$ Oct 14, 2020 at 16:46
  • $\begingroup$ @Andrew I made an edit. Does it help? $\endgroup$
    – Milten
    Oct 14, 2020 at 18:24
  • $\begingroup$ $\sum_{n=1}^x\sum_{m=1}^n = \sum_{m=1}^x\sum_{n=m}^x$ was the problem, for example, let $n=x=2,$ and consider an arbitrary function $f$, then, $\sum_{n=1}^{x=2}\sum_{m=1}^{n=2} f(m)= \underbrace{\sum_{m=1}^{x=1} f(m)+ (\sum_{m=1}^{x=2} f(m))}_{\text{Total} \; x=2 \; \text{times, in other words,}\sum_{n=1}^{x=2}}= f(1)+ (f(1) + f(2))$, but, $\sum_{m=1}^{x=2}\sum_{n=m}^{x=2} f(n)=\sum_{n=1}^{x=2} f(n)+\sum_{n=2}^{x=2} f(n)$ $=f(1)+ f(2) + f(2)$, thus, it looks like, $\sum_{n=1}^x\sum_{m=1}^n \neq \sum_{m=1}^x\sum_{n=m}^x$ , where am i making mistake? $\endgroup$ Oct 14, 2020 at 21:57
  • $\begingroup$ Also, can you mention any book where this kind of sum ($\sum_{n=1}^x\sum_{m=1}^n = \sum_{m=1}^x\sum_{n=m}^x$) has been described, plz? What is the name of this kind of sum, or how it is called? $\endgroup$ Oct 14, 2020 at 22:15
  • $\begingroup$ I think you are making more than one mistake. But most importantly, you can't set $n=x=2$, because only $x$ is a variable you can set. $n$ is a dummy variable, some would call it. So fx with $x=2$, we get: $\sum_{n=1}^2\sum_{m=1}^nf(m) = \sum_{m=1}^1f(m) + \sum_{m=1}^2f(m) = f(1) + (f(1) + f(2))$. And $\sum_{m=1}^2\sum_{n=m}^2f(m) = \sum_{n=1}^2f(1) + \sum_{n=2}^2f(2) = (f(1) + f(1)) + f(2)$. $\endgroup$
    – Milten
    Oct 15, 2020 at 6:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .