### Is the Riemann hypothesis true?

In 2001, Jeffrey Lagarias proved that the Riemann hypothesis is equivalent to the following statement (proof here):

`\sigma(n) <= \H_n + \ln(\H_n)*\exp(\H_n)`

with strict inequality for `n > 1`, where `\sigma(n)` is the sum of the positive divisors of `n`.

In 1913, Grönwall showed that the asymptotic growth rate of the sigma function can be expressed by:

`\lim_{n to \infty}\frac{\sigma(n)}{\n \ln\ln n} = \exp(\gamma)`

where

Relying on this two theorems, we can show that:

`lim_{n to \infty}\frac{\exp(\gamma) * n \ln \ln n}{\H_n + \ln(\H_n) * \exp(\H_n)} = 1`

with strict inequality for each `1 < n < \infty` (see Wolfram|Alpha):

`\exp(\gamma) * n \ln \ln n < \H_n + \ln(\H_n) * \exp(\H_n)`

If the Riemann hypothesis is true, then for each `n ≥ 5041`:

`\sigma(n) <= \exp(\gamma) * n \ln \ln n`

By using the usual definition of the `\gamma` constant:

`\gamma = \lim_{n to \infty}(\H_n - \ln n)`

we can reformulate the result as (see Wolfram|Alpha):

…

`\sigma(n) <= \H_n + \ln(\H_n)*\exp(\H_n)`

with strict inequality for `n > 1`, where `\sigma(n)` is the sum of the positive divisors of `n`.

In 1913, Grönwall showed that the asymptotic growth rate of the sigma function can be expressed by:

`\lim_{n to \infty}\frac{\sigma(n)}{\n \ln\ln n} = \exp(\gamma)`

where

**lim**is the limit superior.Relying on this two theorems, we can show that:

`lim_{n to \infty}\frac{\exp(\gamma) * n \ln \ln n}{\H_n + \ln(\H_n) * \exp(\H_n)} = 1`

with strict inequality for each `1 < n < \infty` (see Wolfram|Alpha):

`\exp(\gamma) * n \ln \ln n < \H_n + \ln(\H_n) * \exp(\H_n)`

If the Riemann hypothesis is true, then for each `n ≥ 5041`:

`\sigma(n) <= \exp(\gamma) * n \ln \ln n`

By using the usual definition of the `\gamma` constant:

`\gamma = \lim_{n to \infty}(\H_n - \ln n)`

we can reformulate the result as (see Wolfram|Alpha):

…