Computing the bound of $d(n)$ the divisor counting "function" of $n$

Question

Terrence Tao gives the following formula for the bound of the divisor $$d(n)\leq \exp\left(O\left({\log n \above 1.5pt \log \log n}\right)\right)$$

How can I use this formula to explicitly compute a bound? This question is part understanding Big-O notation. For example if $n$ is a prime say $17$. Naively I am computing $\exp\left(O\left(\frac{\log 17}{ \log \log 17}\right)\right)=\exp\left(O\left(13.66\right)\right)$. I am not sure if my steps are correct or the next calculation to make? I am assuming the log is base 10?

Answer 1

There is also an effective version, due to Robin and Nicolas. See also Nicolas Survey. $$ d(n) \leq n^{\left( \frac{1.0660186782977...}{ \log \log n} \right)} $$ with the constant $1.066..$ chosen to give equality at $n = 6983776800$ and $d(n) = 2304.$ This is superb for larger numbers. For smaller numbers, better to use $$ d(n) \leq \sqrt {3n} $$ or $$ d(n) \leq 48 \; \sqrt[3] {\frac{n}{2520}} $$

Note $$ \frac{1.0660186782977...}{ \log 2} \approx 1.53793986 $$

Answer 2

$f(n)=O(g(n))$ as $n\to \infty$ means that there exists $k>0$ such that $|f(n))\leq |g(n)|$ for all sufficiently large $n.$ It says nothing about the comparison of $f(n)$ to $g(n)$ for any particular $n.$

We have $$d(n)=\exp \left(O\left(\frac {\log n}{ \log \log n}\right)\right )\iff \log d(n)=O\left( \frac { \log n}{\log n \log n}\right).$$ Since $\log d(n)\geq 0,$ and since $(\log n)/\log \log n>0$ for all but finitely many $n\in \mathbb N,$ this means that there exists $k>0$ such that for all sufficiently large $n$ we have $$\log d(n)\leq k \left(\frac {\log n}{ \log \log n}\right)=\log (n^{k/\log \log n}).$$ One (obvious) implication of this is that for any $j>0$ we have $$\lim_{n\to \infty}\frac {d(n)}{n^j}=0.$$