On $\sigma(n)<n+\frac{n}{2}\log\frac{n+1}{n-1}+\frac{n}{4} \left(\sum_{\substack{d|n,1<d<n}}\log\frac{(n+d)(n+1)}{(n-d)(n-1)}\right)$

Question

I've derived for $n>1$ and $\sigma(n)$ the sum of divisor function $\sum_{d|n}d$ the following inequality $$\sigma(n)<n+\frac{n}{2}\log\frac{n+1}{n-1}+\frac{n}{4} \left(\sum_{\substack{d|n,1<d<n}}\log\frac{(n+d)(n+1)}{(n-d)(n-1)}\right).$$

For the proof, if it isn't wrong, I've used an inequality derived in [1]. Currently I don't made computations with my computer, and I want to know it previous inequality is good in the sense that $RHS/\sigma(n)$ is relatively small.

Question. Can you tell me, in words if the inequality it seems sharper or not, the computational behaviour (with a summary table or a graphic using a computer) of $RHS/\sigma(n)$ for $1<n\leq N$ and $N$ sufficiently large. Of course you can provide us a summary of your results. If you find a counterexample please tell me to discard this inequality. Optionally if you know how estimate this ratio, from a theoretical viewpoint, please tell me. Thanks in advance.

References:

Klambauer, On a Property of $x^ne^{-x}$, The American Mathematical MONTHLY 1988, page 551.

Answer 1

Roughly speaking, your result is that $$ \sigma(n) \ll n \log n,$$ where I use $\ll$ as a sort of big Oh notation. It's actually known that $$ \sigma(n) \ll n \log \log n.$$ The difference between what you've shown and what is known is the difference between $\log n$ and $\log \log n$. In short, the second is much, much smaller. For instance, $\log 10^6 \approx 13.81$ and $\log \log 10^6 \approx 2.62.$ And this difference becomes more pronounced as $n$ gets large, even though it happens slowly.

As $n$ gets large, the ratio between your error and the known error is unboundedly large. But to be fair, the rate of divergence is quite slow.

Answer 2

It isn't clear to me exactly what you are asking the community to provide, but you can compare this to previously established bounds to get a sense for how well you are doing. In particular, a mathematician named Robin showed that $$ \sigma(n) < e^{\gamma}n\log(\log(n)) + \frac{.65n}{\log(\log(n))} $$ holds for all $n\geq 3$. The constant $.65$ isn't quite right, but I don't have my references handy (maybe someone can comment/edit one in), but already this beats your bound for large $n$. So you do not have a "sharp" inequality, but it may have merits in that the proof might be easier than Robin's result. You will probably have to do the computational aspects of your problem yourself. I recommend using Python or Sage.