### 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 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):

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

This implies that the following statement is true for n to \infty:

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

However, this result does not prove Lagarias' inequality for each 1 < n < \infty, which leaves the Riemann hypothesis still an open-problem.