Prove that ## \sum_{n\leq x}\frac{\phi(n)}{\sqrt{n}}=Cx^{3/2}+.... ##.

[Math100]

Homework Statement

Prove that, for $x\geq 2$, $\sum_{n\leq x}\frac{\phi(n)}{\sqrt{n}}=Cx^{3\2}+O(x^{1/2}\log {x})$, where $C$ is a constant that you should determine.

Relevant Equations

If $\alpha\leq 1$ and $x\geq 2$, then $\sum_{n\leq x}\frac{\phi(n)}{n^{\alpha}}=\frac{x^{2-\alpha}}{2-\alpha}\frac{1}{\zeta(2)}+O(x^{1-\alpha}\log {x})$.

Proof:

Let $x\geq 2$.
Then $\sum_{n\leq x}\frac{\phi(n)}{n^{\alpha}}=\sum_{d\leq x}\frac{\mu(d)}{d^{\alpha}}\sum_{q\leq \frac{x}{d}}\frac{1}{q^{\alpha-1}}$.
Observe that
\begin{align*}
&\sum_{n\leq x}\frac{\phi(n)}{\sqrt{n}}=\sum_{d\leq x}\frac{\mu(d)}{\sqrt{d}}\sum_{q\leq \frac{x}{d}}\sqrt{q}\\
&=\sum_{d\leq x}\frac{\mu(d)}{\sqrt{d}}(\frac{2}{3}(x/d)^{3/2}+O(\sqrt{x/d}))\\
&=\frac{2}{3}x^{3/2}\sum_{d\leq x}\frac{\mu(d)}{d^{2}}+O(\sqrt{x}\sum_{d\leq x}\frac{\mu(d)}{d})\\
&=\frac{2}{3}x^{3/2}\frac{1}{\zeta(2)}-\frac{2}{3}x^{3/2}\sum_{d>x}\frac{\mu(d)}{d^{2}}+O(\sqrt{x}\log {x})\\
&=\frac{2}{3}x^{3/2}\frac{1}{\zeta(2)}+O(\sqrt{x})+O(\sqrt{x}\log {x})\\
&=\frac{2}{3}x^{3/2}\frac{6}{\pi^{2}}+O(\sqrt{x}+\sqrt{x}\log {x})\\
&=\frac{4}{\pi^{2}}x^{3/2}+O(x^{1/2}\log {x}).\\
\end{align*}
Therefore, $\sum_{n\leq x}\frac{\phi(n)}{\sqrt{n}}=Cx^{3/2}+O(x^{1/2}\log {x})$, where $C=\frac{4}{\pi^{2}}$ is a constant.

 

[Math100]

After factoring out the big-O notation, how does $\sqrt{x}+\sqrt{x}\log {x}=x^{1/2}\log {x}$?

 

[fresh_42]

After factoring out the big-O notation, how does $\sqrt{x}+\sqrt{x}\log {x}=x^{1/2}\log {x}$?

Those O-notations are called Landau symbols. Here is a list
https://en.wikipedia.org/wiki/Big_O_notation#Family_of_Bachmann–Landau_notations
and explanations.

 

[WWGD]

After factoring out the big-O notation, how does $\sqrt{x}+\sqrt{x}\log {x}=x^{1/2}\log {x}$?

I assume it's factored as $O(\sqrt(1 +logx)=O(x^{1/2}(1+log x))$, and the $1$ by itself is absorbed into $log x$.

 

[fresh_42]

I assume it's factored as $O(\sqrt(1 +logx)=O(x^{1/2}(1+log x))$, and the $1$ by itself is absorbed into $log x$.

It is easier to consider it as $0\cdot \sqrt{x}\log(x)< \sqrt{x}+\sqrt{x}\log(x) < 2\sqrt{x}\log(x).$