The summatory function of the reciprocal of the totient is :S(n) := \sum _{k=1}^{n}{\frac {1}{\varphi (k)}}.
Edmund Landau showed in 1900 that this function has the asymptotic behavior :S(n) \sim A (\gamma+\log n)+ B +O\left(\frac{\log n} n\right), where is the
Euler–Mascheroni constant, :A = \sum_{k=1}^\infty \frac{\mu (k)^2}{k \varphi(k)} = \frac{\zeta(2)\zeta(3)}{\zeta(6)} = \prod_{p\in\mathbb{P}} \left(1+\frac 1 {p(p-1)} \right), and :B = \sum_{k=1}^{\infty} \frac{\mu (k)^2\log k}{k \,\varphi(k)} = A \, \prod _{p\in\mathbb{P}}\left(\frac {\log p}{p^2-p+1}\right). The constant is sometimes known as '''Landau's totient constant'''. The sum \textstyle \sum _{k=1}^\infty 1 / (k \; \varphi (k)) converges to :\sum _{k=1}^\infty \frac 1 {k\varphi (k)} = \zeta(2) \prod_{p\in\mathbb{P}} \left(1 + \frac 1 {p^2(p-1)}\right) =2.20386\ldots. In this case, the product over the primes in the right side is a constant known as the
totient summatory constant, and its value is :\prod_{p\in\mathbb{P}} \left(1+\frac 1 {p^2(p-1)} \right) = 1.339784\ldots. == See also ==