A bound for $\log\ n>3k\log\log n $.

Question

Suppose $k$ is natural number. I want to find the best bound for $\log\ n>3k\log\log n $.

That is, if we have $n>A$ then $\log\ n>3k\log\log n$, where $A$ is term according to $k$.

Answer 1

$\ln n/\ln\ln n$ is an increasing function for $n > e^e$. So the bound $A$ will satisfy $\ln A = 3k\ln\ln A$ and $A > e^e$. Let $\alpha = \ln\ln A$. Then we have $$ e^\alpha = 3k \alpha\Longrightarrow -\alpha e^{-\alpha} = -\frac{1}{3k}\Longrightarrow \alpha = -W_{-1}\left(- \frac{1}{3k}\right), $$ where $W_{-1}$ is a branch Lambert $W$ function, defined implicitly by $W_{-1}(z)\exp W_{-1}(z) = z$ and $-W_{-1}(z) > e$ for $-1 <z < 0$. This gives $$ A = \exp\exp\left[-W_{-1}\left(- \frac{1}{3k}\right)\right] = \exp\left[-3kW_{-1}\left(- \frac{1}{3k}\right)\right] = \left[-3kW_{-1}\left(- \frac{1}{3k}\right)\right]^{3k} $$ Now, for $z$ close to zero, $W_{-1}(-z)\sim -\ln(z)$. So for large $k$, we have $A\sim[3k\ln(3k)]^{3k}$. This extraordinarly fast growth shows just how slowly $\ln n$ goes to infinity.