I am trying to calculate the $k$ value in this equation:
$\dfrac1{n^c} \le \left(1 - \dfrac2{n(n-1)} \right)^k$
by using the logarithm, I am getting for $k$:
$\log_{1- 2/n(n-1)} n^{-c} \le k$
is that correct? And ist there any way to simplify the result of $k$?
If you take logs, $\dfrac1{n^c} \le \left(1 - \dfrac2{n(n-1)} \right)^k $ becomes $-c\log(n) \le k \log\left(1 - \dfrac2{n(n-1)} \right) $.
Since $\log\left(1 - \dfrac2{n(n-1)} \right) < 0$, dividing by it reverses the sign of the inequality, so we get $\frac{-c\log(n)}{\log\left(1 - \dfrac2{n(n-1)} \right)} \ge k $.
If $n$ is large compared with $k$, the right side is about $1 - \dfrac{2k}{n(n-1)} $, so we get, as an approximation, $\dfrac1{n^c} \le 1 - \dfrac{2k}{n(n-1)} $ or $k \le \frac12 n(n-1)(1-n^{-c}) $.