Answer
$1.195$
Work Step by Step
Consider $\int_n^\infty \dfrac{1}{x^3}dx \lt 0.01$
and $ \lim\limits_{k \to \infty} (\dfrac{-1}{2k^2}+\dfrac{1}{2n^2})=\dfrac{1}{2n^2} \lt 0.01$
Also, $n >\sqrt {50} \approx 7.071 \implies n \geq 8$
We will take the help of Recursion mode.
Then, $S\approx s_8= \Sigma_{n=1}^8 \dfrac{1}{n^3}$
and $\Sigma_{n=1}^8 \dfrac{1}{n^3} \approx 1.195$
