Science:Math Exam Resources/Courses/MATH101/April 2013/Question 10 (b)/Solution 1

By part (a),

$\displaystyle {\begin{aligned}\int _{0}^{1/4}{\frac {1}{1+x^{3}}}\ dx&=\left.\left[\left(\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}x^{3n+1}\right)+C\right]\right|_{0}^{1/4}\\&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}\left({\frac {1}{4}}\right)^{3n+1}\end{aligned}}$

This is an alternating sum with

$\displaystyle b_{n}={\frac {1}{3n+1}}\left({\frac {1}{4}}\right)^{3n+1}.$

Clearly $\displaystyle b_{n+1}\leq b_{n}$ and $\displaystyle \lim _{n\to \infty }b_{n}=0$. Then by Alternating Series Remainder Theorem, the error in the approximation is bounded as follows:

$\displaystyle |{\text{error}}|\leq b_{n+1}$

Then we are guaranteed the error is less than $\displaystyle 10^{-5}$ if $\displaystyle b_{n+1}\leq 10^{-5}$, that is, we must solve

$\displaystyle {\frac {1}{3(n+1)+1}}\left({\frac {1}{4}}\right)^{3(n+1)+1}\leq 10^{-5}={\frac {1}{10^{5}}}.$

For $\displaystyle n+1=1$,

$\displaystyle {\frac {1}{3+1}}\left({\frac {1}{4}}\right)^{3+1}={\frac {1}{4^{5}}}>{\frac {1}{10^{5}}}$

so we do not have enough terms yet. For $\displaystyle n+1=2$,

$\displaystyle {\frac {1}{3(2)+1}}\left({\frac {1}{4}}\right)^{3(2)+1}\leq {\frac {1}{7}}{\frac {1}{16000}}={\frac {1}{112000}}<{\frac {1}{10^{5}}}$

where we have used $\displaystyle 4^{7}=2^{14}=2^{10}\cdot 16=1024\cdot 16\geq 1000\cdot 16=16000$ to approximate. Thus, $\displaystyle n+1=2$ is sufficient. Since the series starts at 0, we need 2 terms (the $\displaystyle n=0$ and $\displaystyle n=1$ term, specifically).