Answer
$0.57$
Work Step by Step
Since, we have $\int_n^\infty \dfrac{1}{x^2+4}dx \lt 0.1$
or, $\lim\limits_{a \to \infty} (\dfrac{1}{2} tan^{-1} (a/2) -\dfrac{1}{2} tan^{-1} (n/2))\lt 0.1$
or, $n \gt 2 \tan (\pi/2-0.2) \approx 9.867 $
or, $n \geq 10$
Need to use Recursion mode.
we get:
$S\approx s_{10}= \Sigma_{n=1}^{10} \dfrac{1}{n^2+4} \approx 0.57$
