The sum of the first $\displaystyle n$ terms of a series is given by the expression $\displaystyle (1-3^{-2n})$.

Find the least value of $\displaystyle k$ such that the sum of the terms from the $\displaystyle k^{th}$ term is less than $\displaystyle \frac{1}{3000}$

$\displaystyle 1-3^{2k} < \frac{1}{3000}$

$\displaystyle -3^{-2k} < -\frac{2999}{3000}$

$\displaystyle 3^{-2k} > \frac{2999}{3000}$

$\displaystyle -2k < \frac{lg\frac{2999}{3000}}{lg3}$

$\displaystyle k > 0.0001517$

but ans is 5