An upper bound for the number of divisors

Question

I've come across this problem in Murty & Carmen, exercise 1.5.3: show that there is a constant $c$ such that $d(n)=O(\exp(\frac{c\log n}{\log\log n})))$ where $d(n)$ is the number of divisors of $n$

I've gotten most of the way through and have also looked through this article https://terrytao.wordpress.com/2008/09/23/the-divisor-bound/, but I don't understand this one step in their explanation:

$(6)\le O(1/\varepsilon)^{\exp(1/\varepsilon)} = \exp\exp(O(1/\varepsilon))$

Could anyone provide some justification as to why this would be true? I've tried playing around with the equality, but I've been having no luck. Thanks in advance!

Answer 1

By unique factorization theorem, we can write $n$ into

$$ n=p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}q_1^{s_1}q_2^{s_2}\cdots q_m^{s_m} $$ where $p_i$ denotes some prime $\le t$, and $q_j$ denote some prime $>t$. Thus, we have

$$ d(n)=\prod_i(r_i+1)\prod_j(s_j+1) $$ By definition, we know that $k\le t$ and $r_i<\log_2 n$, so we have

\begin{aligned} d(n) &\le\left(1+\log_2n\right)^k\prod_j(s_j+1)\le(1+\log_2n)^t\prod_j2^{s_j} \\ &=(1+\log_2n)^t\prod_j q_j^{s_j\log2/\log q_j}<(1+\log_2n)^t\prod_jq_j^{s_j\log2/\log t} \\ &\le(1+\log_2n)^tn^{\log2/\log t}=\exp\{t\log(1+\log_2n)+\log2\log n/\log t\} \\ &\le2^{\log n/\log t+O(t\log\log n)} \end{aligned}

Finally, setting $t=\log n/(\log\log n)^3$ gives $d(n)<2^{(1+\varepsilon)\log n/\log\log n}$ for large $n$.

For OP's question on the specific inequality, let $A_1,A_2,\dots$ denote the O constants then \begin{aligned} (A_1\varepsilon^{-1})^{\exp(\varepsilon^{-1})} &=\exp[\exp(\varepsilon^{-1})\log(A_1\varepsilon^{-1})] \\ &<\exp[\exp(\varepsilon^{-1})\exp(A_2\varepsilon^{-1})] \\ &=\exp\{\exp[(1+A_2)\varepsilon^{-1}]\}=\exp\exp O(\varepsilon^{-1}) \end{aligned}