Answer
$1.195$
Work Step by Step
Since, we have $\int_n^\infty \dfrac{1}{x^3}dx \lt 0.01$
or, $ \lim\limits_{a \to \infty} (\dfrac{-1}{2a^2}+\dfrac{1}{2n^2})=\dfrac{1}{2n^2}\lt 0.01$
or, $n \sqrt {50} \approx 7.071 $
or, $n \geq 8$
Need to use Recursion mode.
we get:
$S\approx s_8= \Sigma_{n=1}^8 \dfrac{1}{n^3} \approx 1.195$
