Estimate of $\prod_{d\mid n}(d+1)$.

Question

I would like to ask if there is any cool estimate of $\prod_{d\mid n}(d+1)$.
I know that $\prod_{d\mid n}d=n^{\tau(n)/2}$ ($\tau(n)$ is the number of divisors of $n$) so we have the trivial estimate $\prod_{d\mid n}(d+1)>n^{\tau(n)/2}$.
The question is, is there any hope for something better? The answer seems to be yes, since for big $\tau(n)$ the difference $\prod_{d\mid n}(d+1)-\prod_{d\mid n}d$ seems to be large, but I was unable to find something precise.

Answer 1

If you're looking for asymptotic formulae on the average $\sum_{n<N}\prod_{d|n}(1+d)$, then I do not think that you will find any. Let $F(n)=\prod_{d|n}(1+d)$ and let us assume that we have indeed found representation in the form

$$\frac{1}{N}\sum_{n<N}F(n)=M(N)+R(N)$$

where $R(N)$ is some "small" error term and $M(N)$ is a main term. The weakest possible equation of this form would be $R(N)=o\left(M(N)\right)$, i.e

$$\frac{1}{N}\sum_{n<N}F(n)\sim M(N)$$

This would imply that

\begin{align*} \left|\frac{1}{N}\sum_{n<N}\log(F(n))-\log(M(N))\right|&=\left|\frac{1}{N}\sum_{n<N}\log\left(\frac{F(n)}{M(n)}\right)\right|\\ &=\left|\frac{1}{N}\sum_{n<N}\log\left(1+\frac{F(n)-M(n)}{M(n)}\right)\right|\\ &\sim \left|\frac{1}{N}\sum_{n<N}\frac{F(n)-M(n)}{M(n)}\right|\\ &=\frac{R(n)}{M(n)}\\ &=o(1) \end{align*}

The sums $\sum_{d|n}\log(1+d)$, however, do not go to zero and are not perfectly behaved and so it would be unreasonable to expect a formula in the form

$$\frac{1}{N}\sum_{n<N}\sum_{d|n}\log(1+d)=M(N)+o(1)$$

we can, however, examine

\begin{align*} \frac{1}{N}\sum_{n<N}\sum_{d|n}\log(1+d)&=\frac{1}{N}\sum_{n<N}\sum_{d|n}\log\left(1+\frac{1}{d}\right)+\frac{1}{N}\sum_{n<N}\sum_{d|n}\log\left(d\right) \end{align*}

We get now that

\begin{align*} \frac{1}{N}\sum_{n<N}\sum_{d|n}\log\left(1+\frac{1}{d}\right)&=\frac{1}{N}\sum_{n<N}\sum_{d|n}\frac{1}{d}+O\left(\frac{1}{N}\sum_{n<N}\sum_{d|n}\frac{1}{d^2}\right)\\ &=\frac{1}{N}\sum_{d<N}^{ }\sum_{n<\frac{N}{d}}^{ }\frac{1}{d}+O\left(\frac{1}{N}\sum_{d<N}^{ }\sum_{n<\frac{N}{d}}^{ }\frac{1}{d^{2}}\right)\\ &=\sum_{d<N}^{ }\frac{1}{d^{2}}+O\left(\frac{\log\left(N\right)}{N}\right)+O\left(\sum_{d<N}^{ }\frac{1}{d^{3}}\right)\\ &=O\left(1\right) \end{align*}

and so

$$\frac{1}{N}\sum_{n<N}\sum_{d|n}\log(1+d)=\frac{1}{N}\sum_{n<N}\log(n)\frac{\tau(n)}{2}+O(1)$$

you can now estimate this RHS with whatever classical results you want on the distribution of $\tau(n)$. The best results have a worse than $O(1)$ error (obviously) and so you'll get a claim of equal strength to what is state of the art.