Science:Math Exam Resources/Courses/MATH101/April 2016/Question 11 (b)/Solution 2

An alternative method is to use the Integral test to approximate the error.

First we note that

$|S-S_{5}|=|\sum _{n=6}^{\infty }(-1)^{n-1}24n^{2}e^{-n^{3}}|\leq \sum _{n=6}^{\infty }|(-1)^{n-1}24n^{2}e^{-n^{3}}|=\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}.$

so it is sufficient to find an upper bound for $\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}.$

Now if $a_{n}=24n^{2}e^{-n^{3}}$, then we showed that $a_{n}$ is positive and decreasing, so we can apply the integral test to estimate $\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}$ with $\int _{5}^{\infty }24x^{2}e^{-x^{3}}\,dx$.

This integral represents the area under the graph of $f(x)=24x^{2}e^{-x^{3}}$

But $\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}$ represents the sum of the area of rectangles starting at $x=5$ with width $1$, which is in fact the Reimann sum of $f(x)$ with right endpoints on each subinterval $[5,6],[6,7],...$, so it underestimates the actual area under the graph, therefore;

$\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}\leq \int _{5}^{\infty }24x^{2}e^{-x^{3}}\,dx=8e^{-5^{3}}.$

$\color {blue}|S-S_{5}|\leq 8e^{-5^{3}}.$