Answer
$\frac{1}{\sqrt[10] e}=e^{-0.1}\approx 0.90483$
Work Step by Step
Given:$\frac{1}{\sqrt[10] e}=e^{-0.1}$
The series for $e^{x}=\Sigma_{n=0}^{\infty}\frac{x^{n}}{n!}$
Put $x=-0.1$
$e^{x}=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(0.1)^{n}}{n!}$
Here, $a_{n}=(-1)^{n}\frac{(0.1)^{n}}{n!}$
So, $s_{2}=a_{0}+a_{1}+a_{2}+a_{3} \approx 0.90483$
Thus, $\frac{1}{\sqrt[10] e}=e^{-0.1}\approx 0.90483$
