On $\frac{\sigma(n+1)}{p_{n+1}}-\frac{\sigma(n)}{p_{n}}$, where $p_k$ is the $kth$ prime number and $\sigma(k)$ the sum of divisors function

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Few minutes ago while I was reading the Wikipedia's article for Andrica's conjecture I wondered about this variation, where $$\sigma(m)=\sum_{d\mid m}d,$$ is the sum of divisor function:

Conjecture. For each integer $n>1$ one has that $$ \left| \frac{\sigma(n+1)}{p_{n+1}}-\frac{\sigma(n)}{p_{n}} \right| <\frac{1}{2}.$$

I don't know if such was in the literature or it is easy to deduce. Obviously I know that here $\sigma(n)>n$, and more about the size of this arithmetic function, but the sum of divisor function is erratic. Also I know different versions of the Prime Number Theorem.

With Wolfram Alpha online calculator one can see that it isn't easy to find counterexamples

plot sigma(n+1)/Prime(n+1)-sigma(n)/Prime(n), for n=1 to 1000000

I don't know if such behaviour is due that this conjecture is good, or is due that it is an obvious proposition (that can be easily deduced as true).

Question. If this conjecture is well-known please refer the literature. In other case, do you know how to start to dilucidate if it is true, or false? Many thanks.

Motivation. As remark, if one writes the inequality as $ \left| \frac{1}{p_{n+1}}\sigma(n+1)-\frac{1}{p_{n}}\sigma(n) \right| <\frac{1}{2}$, then one can thinks in this condition (and maybe these words here have no mathematical meaning) as the difference of two integrals (an average or smoothing of these), if one thinks in the sum of divisors function as an integral, and in the $nth$ prime number as a lenght, or volume of primes at $n$.

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By the Prime Number Theorem, $$ p_n \sim n \log n. $$

By Gronwall's theorem, $$ \limsup\limits_{n\to\infty} {\sigma(n) \over n\log\log n} = e^\gamma\approx1.781. $$

It follows from the PNT and Gronwall's theorem that there are at most finitely many exceptions to your conjecture.

Moreover, Robin's theorem states that the Riemann hypothesis is equivalent to $$ \sigma(n) < e^\gamma n \log\log n \quad\mbox{ for all } n\ge5041. \tag{1} $$ Therefore, if the Riemann hypothesis is true, your conjecture is also true. This is not difficult to verify by computer using inequality $(1)$ in combination with $$ p_n \ge n \left(\log n + \log\log n - 1 + {\log\log n - 2.1\over \log n}\right) \qquad\mbox{ for } n\ge 3 \tag{2} $$ (see arXiv:1002.0442).

Edit: Regardless of the truth of the Riemann hypothesis, Robin also proved that $$ {\sigma(n)\over n\log\log n} \le e^\gamma + {0.6482\ldots\over(\log\log n)^2} \quad\mbox{ for all } n\ge3, $$ with equality for $n=12$. This result, in combination with the above bound $(2)$ for $p_n$, implies that your conjecture is true, regardless of the Riemann hypothesis.