Question on the zeta and sigma functions

Question

EDIT: The observation described here was due to a bug in the code used to generate the data (as commented by Lucia), which renders this question irrelevant. The answer, however, is worth reading.

The following relation is quite easy to derive $$(\sum_{i=1}^n \frac{1}{i^R})^2 = \sum_{i=1}^n \frac{\sigma_1(i)}{i^R}+\epsilon$$ With $\lim_{n\to\infty}\epsilon = 0$

For any fixed $n$ one can describe the error term as a function of $R$. Ratios of consecutive values of said function seem to be independent from $n$. $${\epsilon_{R-1}}/{\epsilon_R} = \Omega_R \approx 4$$ $\Omega_R$ appears at first to decrease monotonically towards convergence, but some values are greater than expected, hinting that it diverges. Below is a link to a python script which can list some of its values.

My question is: what could possibly account for the irregular behavior of $\Omega$? is its growth bounded?

First few error ratios: https://repl.it/FY1v/3

Answer 1

My original response was wrong in an embarrasing way, so let me fix it. As Lucia remarked, the OP's observation was due to a programming error, and the general picture is as follows. We have $$\epsilon_R(n)=\left(\sum_{m=1}^n \frac{1}{m^R}\right)^2-\sum_{m=1}^n \frac{d(m)}{m^R}=\sum_{m=n+1}^{n^2}\frac{c(m)}{m^R},$$ where $c(m)$ counts the number of representations $m=m_1m_2$ with $m_1,m_2\leq n$. Now let $n\geq 2$ be fixed, and let $R\to\infty$. Then we get $$\epsilon_R(n)=\begin{cases}\frac{c(n+1)+o(1)}{(n+1)^R}&\text{when $n+1$ is composite};\\\frac{c(n+2)+o(1)}{(n+2)^R}&\text{when $n+1$ is a prime.}\end{cases}$$ In particular, $$ \lim_{R\to\infty}\frac{\epsilon_{R-1}(n)}{\epsilon_R(n)}=\begin{cases}n+1&\text{when $n+1$ is composite};\\n+2&\text{when $n+1$ is a prime.}\end{cases}$$

Added. Here are some numerical experiments done with SAGE that confirm the above: $$\frac{\epsilon_{499}(99)}{\epsilon_{500}(99)}\approx\frac{7.00030668509\times 10^{-998}}{7.00030067129\times 10^{-1000}}\approx 100.000085908$$ $$\frac{\epsilon_{499}(100)}{\epsilon_{500}(100)}\approx\frac{3.06685088585\times 10^{-1002}}{3.00671292947\times 10^{-1004}}\approx 102.000122985$$ SAGE code used:

R=500;n=100;
A=sum([RealField(5000)(1/m^(R-1)) for m in range(1,n+1)])^2-sum([RealField(5000)(number_of_divisors(m)/m^(R-1)) for m in range(1,n+1)])
B=sum([RealField(5000)(1/m^R) for m in range(1,n+1)])^2-sum([RealField(5000)(number_of_divisors(m)/m^R) for m in range(1,n+1)])
A/B