Are these equalities wrong $\sum_{n\le x}\mu(n)o(\frac xn)=o(x\sum_{n\le x}\frac 1n)=o (x\log x)$?

I simply found an asymptotic relation for $$\sum_{n\le x}\mu(n)o(\frac xn)$$ like below:

$$\sum_{n\le x}\mu(n)o(\frac xn)=o(x\sum_{n\le x}\frac 1n)=o (x\log x)$$

But Basil Gordon, an American mathematician, in his thesis found that like below:

$$\sum_{n\le x}\mu (n)o (\frac xn)=\sum_{n\le\frac x{\log x}}\mu (n)o (\frac xn)+\sum_{\frac x{\log x}\le n \le x}\mu (n)O (\frac xn)$$

Then since $$|\mu (n)|\le 1$$

$$o (\sum_{n\le\frac x{\log x}}\frac xn)=o (x\log x)$$

and

$$O (\sum_{\frac x {\log x}\le n\le x}\frac xn)=O (x\log x-x\log\frac x{\log x})=O (x\log\log x)=o (x\log x)$$

Could anyone explain for me why Basil Gordon has done that like this? Is my answer wrong?

• Big-oh of $x\log\log x$ is sharper than little-oh of $x\log x$. – Gerry Myerson Jun 28 at 12:51
• @Gerry Myerson My question is when we can arrive to $o (x\log x)$ by a short proof, why Basil Gordon arrived to $o(x\log x)$ by a long proof? – user678879 Jun 28 at 17:36
• How do you justify $\sum_{n\le x}\mu(n)o(\frac xn)=o(x\sum_{n\le x}\frac1n)$ – reuns Jun 28 at 17:49
• @reuns Since $|\mu (n)|\le 1$ it follows from $o (f)+o (g)=o (f+g)$. – user678879 Jun 28 at 17:52
• No. $\sum_{n \le x} \mu(n) o(1/n^2) = O(\sum_{n \le x} 1/n^2)=O(1)$ not $o(1)$. – reuns Jun 28 at 17:53

Claim(?): Let $$f(n,x)$$ be a function in $$n$$ and $$x$$. Suppose $$\lim \frac{f(n,x)}{x/n} = 0$$. Then we have $$\lim \frac{\sum_{n\leq x} f(n,x)}{\sum_{n \leq x} x/n} = 0$$.
Is this claim true? It depends on context (so I intentionally omit how the limit goes). What you wrote in your question, you interpreted small o notation by $$\lim_{n\leq x, x\rightarrow\infty} f(n,x) = 0$$ uniformly in $$n\leq x$$.
I haven't looked into his thesis, but based on what was written, I assume what the author meant by the formula $$o(x/n)$$ is any function of the form $$\phi(x/n)$$ where $$\phi$$ is a function which satisfies $$\lim_{u\rightarrow \infty} \phi(u) / u = 0$$. (or, $$o(x/n)$$ uniformly as $$x/n \rightarrow \infty$$). What makes you more confusing is that, in the last formula $$o(x log x)$$, his limit was taken as $$x \rightarrow \infty$$. So it is indeed the same notation with different meaning from line to line. So, I'll write $$o_{x/n}$$ to denote the former and $$o_x$$ for the latter. Then your interpretation is $$\sum_{n\leq x} o_x \left ( \frac{x}{n} \right ) = x o_x \left (\sum_{n\leq x} \frac{1}{n} \right ) = o_x (x \log x),$$ which is obviously true. I believe the author's intention was $$\sum_{n\leq x} o_{x/n} (x/n) = o_x ( x \log x),$$ which is not automatic(for instance, when $$x/2\leq n\leq x$$, $$x/n$$ is bounded so we don't know if it is $$o_x(1)$$. Thus the author goes in two steps to handle this.
What reuns said I believe is that $$\sum_{n\leq x} o_n (1/n^2) \neq o_x(1)$$. So it all depends on the context after all.