let $\Psi (n,z)=\sum\limits_{\substack{d|n \\ p|d \Rightarrow p>z}}{1\over d}$，I want to find a "z" as small as possible related to n so that $\Psi (n,z)=O(1)$ or $\Psi (n,z)=o(1)$
the below is what I have done:
$$ \begin{aligned} \Psi(n,z) &=\sum\limits_{\substack{d|n \\ p|d \Rightarrow p>z}} {1\over d} \\ &=\prod\limits_{\substack{p^k ||n \\ p>z}} (1+{1\over p}+\dots+{1\over p^k}) \;\;\;(Euler \;product\; formula )\\ & \le \prod\limits_{\substack{p |n \\ p>z}} (1+ {1\over p-1}) \;\;\;(summation \;formula \;of\;geometric\;series ) \end{aligned} $$
Then，take nature logarithm.
We get, $$ \begin{aligned} \log\Psi(n,z) &=\sum\limits_{\substack{p|n \\ p>z}} \log (1+{1\over p-1}) \;\;\; \\ &\le \sum\limits_{\substack{p|n \\ p>z}} {1\over p-1} \;\;\;(a \;inequality \;about\; logarithm)\\ &\le \sum\limits_{\substack{p|n }} {1\over z-1} \\ &\le {1 \over z-1} \log n \end{aligned} $$
so we get $ \begin{aligned} \log\Psi(n,z) &\le {1 \over z-1} \log n \end{aligned}$.
$\begin{aligned} \Psi(n,z) &\le e^{ {1 \over z-1} \log n} \end{aligned}$
If we let $z=\log n +1$, then we have :
$\begin{aligned} \Psi(n,z) &\le e^{ {1 \over z-1} \log n}=e=O(1) \end{aligned}$
so, we get a "z" related to n ,$z=\log n +1$ so that $ \begin{aligned} \Psi(n,z) =O(1) \end{aligned}$
But I feel my result seems too weak, is there any stronger result?
The stronger the better.
Thank you very much !
After thinking for a period of time, I find it seems not particularly difficult to make stronger results.
It perhaps needs one or two small tricks .
Let's review the previous derivation steps
$ \begin{aligned} \log\Psi(n,z) &=\sum\limits_{\substack{p|n \\ p>z}} \log (1+{1\over p-1}) \;\;\; \\ &\le \sum\limits_{\substack{p|n \\ p>z}} {1\over p-1} \;\;\;\\ &\le \sum\limits_{\substack{p|n }} {1\over z-1} \\ &\le {1 \over z-1} \log n \end{aligned} $
Let's focus on $\begin{aligned} \sum\limits_{\substack{p|n \\ p>z}} {1\over p-1} \end{aligned}$ and split it .
we have ,
$\begin{aligned} \sum\limits_{\substack{p|n \\ p>z}} {1\over p-1} = \underbrace{\sum\limits_{\substack{p|n \\ z < p < \log n +1}} {1\over p-1} }_{S_1} +\underbrace{ \sum\limits_{\substack{p|n \\ \log n +1 \le p }} {1\over p-1} }_{S_2} \end{aligned}$
The calculation of $S_2$
Now consider the number of different prime factors in n that are greater than or equal to $ \log n + 1$, and record this number as a
We have ,
$\begin{aligned} \left( \log n+1 \right)^a\le \prod_{1\le r \le a} p_r \le n \end{aligned}$ ,where $p_r,1\le r \le a$,is the prime factor of n that greater than or equal to $ \log n + 1$
So we get , $\begin {aligned} \sum_{ \substack {p|n \\ \log n+1 \le p } } 1=a\le {\log n \over \log (\log n +1)} < {\log n \over \log \log n } \end {aligned}$
$\begin{aligned} S_2&={ \sum\limits_{\substack{p|n \\ \log n +1 \le p }} {1\over p-1} }\\ &\le { \sum\limits_{\substack{p|n \\ \log n +1 \le p }} {1\over \log n} } \\ &< {1\over \log n} { { \log n\over \log \log n} } \;\;\;(\sum_{ \substack {p|n \\ \log n+1 \le p } } 1 < {\log n \over \log \log n } used\; here )\\ &={1 \over \log \log n } \end{aligned}$
The calculation of $S_2$ here is basically the same as the derivation of the description of the question.
This splitting makes the range of z smaller ,So we just need to focus on the smaller range of z .
The calculation of $S_1$
$\begin{aligned} S_1={\sum\limits_{\substack{p|n \\ z < p < \log n +1}} {1\over p-1} } \le {\sum\limits_{\substack{ \\ z < p < \log n +1}} {1\over p-1} } \end{aligned}$
Here, we can easily think of Mertens's second theorem.
mertens's second theorem $ \sum_{p\le x}\frac1p=\log\log x+B_1+\mathcal O\left(1\over\log x\right) \tag1 $where $B_1=\lim_{x\to\infty}\left[\sum_{p\le x}\frac1p-\log\log x\right]$ is the mertens constant.
First, we make a difference between $\begin{aligned} {\sum\limits_{\substack{ \\ z < p < \log n +1}} {1\over p-1} } \end{aligned}$ and $\begin{aligned} {\sum\limits_{\substack{ \\ z < p < \log n +1}} {1\over p} } \end{aligned}$
We have ,
$ \begin{aligned} {\sum\limits_{\substack{ \\ z < p < \log n +1}} \left | {1\over p-1} -{ 1 \over p}\right |} &={\sum\limits_{\substack{ \\ z < p < \log n +1}} \left | {1 \over p\,(p-1)} \right |} \\ & <{\sum\limits_{\substack{ \\ z < p < \log n +1}} \left | {1 \over \,(p-1)^2} \right |} \\ & <{\sum\limits_{\substack{ \\ z < t < \log n +1}} \left | {1 \over \,(t-1)^2} \right |} \\ & \ll { 1 \over z} \end{aligned} $
So
$ \begin{aligned} S_1\le{\sum\limits_{\substack{ \\ z < p < \log n +1}} {1\over p-1} }=\begin{aligned} {\sum\limits_{\substack{ \\ z < p < \log n +1}} {1\over p} }+ \mathcal O \left(\;{1 \over z} \;\right) \end{aligned} \end{aligned} $
Applying mertens second theorem (1) :
We have $\begin{aligned} S_1&\le {\sum\limits_{\substack{ \\ z < p < \log n +1}} {1\over p} }+ \mathcal O \left(\;{1 \over z} \;\right) \\ &=\log \log (\log n+1)-\log \log z+\mathcal O \left(\;{1 \over z} \;\right) +\mathcal O \left(\;{1 \over \log n} \;\right) \end{aligned}$
Consider that $z<\log n+1$,The $\mathcal O$ can be merged.
So we have $\begin{aligned} S_1 &\le\log \log (\log n+1)-\log \log z+\mathcal O \left(\;{1 \over z} \;\right) \\ &= \log \left( {\log(\log n +1) \over \log z } \right) + \mathcal O \left(\;{1 \over z} \;\right) \end{aligned}$
So far, we have completed the calculation of $S_1$ and $S_2$.
Reviewing the previous equation,
$\begin{aligned} \sum\limits_{\substack{p|n \\ p>z}} {1\over p-1} = \underbrace{\sum\limits_{\substack{p|n \\ z < p < \log n +1}} {1\over p-1} }_{S_1} +\underbrace{ \sum\limits_{\substack{p|n \\ \log n +1 \le p }} {1\over p-1} }_{S_2} \end{aligned}$
Let's substitute $ \begin{aligned} S_1 &\le \log \left ( {\log(\log n +1) \over \log z } \right) + \mathcal O \left(\;{1 \over z} \;\right) \end{aligned}$ and $\begin{aligned} S_2 < {1 \over \log \log n} \end{aligned}$ into the equation
We have
$ \begin{aligned} \log\Psi(n,z) &\le \sum\limits_{\substack{p|n \\ p>z}} {1\over p-1} \le \log \left ( {\log(\log n +1) \over \log z } \right) + \mathcal O \left(\;{1 \over z} \;\right) +\mathcal O \left(\;{1 \over \log \log n} \;\right) \end{aligned} $
This time we can get a smaller z .
For example ,
Let $z=\sqrt{\log n+1}$ ,this time we only observe $\begin{aligned} \log \left( {\log(\log n +1) \over \log z } \right) \end{aligned}$
Then,
$\begin{aligned} \log \left ( {\log(\log n +1 \over \log z } \right)&= \log \left ( {\log(\log n +1) \over \log (\sqrt{\log n+1}) } \right)\\ &=\log \left ( {2\log(\log n +1) \over \log (\log n+1) } \right)\\ &=\log2 \end{aligned}$
$\log 2$ is a constant.
Through this derivation, in fact, we can always take a fixed number $\varepsilon$ greater than 0, let $z=\log ^\varepsilon n$, then $\Psi(n,z) =\mathcal O(1)$