Answer
$R_{4}\leq\frac{1}{2e^{16}}$
$0.4049\leq S\leq0.4088$
Work Step by Step
$\Sigma_{n-1}^{\infty}ne^{-n^{2}}$, using $n=4$
Find the partial sum $S_{4}$ by plugging it into a calculator:
$S_{4}\Sigma_{n-1}^{4}ne^{-n^{2}}\approx0.4048813985$
$S_{4}=e^{-1}+2e^{-4}+3e^{-9}+4e^{-16}\approx0.4048813985$
The function $ne^{-n^{2}}$ is positive, continuous, and ultimately decreasing. Therefore we can use the integral test to find the approximation of maximum error for the function.
$R_{4}\leq\int_{4}^{\infty}ne^{-n^{2}}dn$
$R_{4}\leq\lim\limits_{a \to \infty}\int_{4}^{a}ne^{-n^{2}}dn$
$let$ $u=-n^{2}$
$du=-2ndn$
$-\frac{1}{2}du=ndn$
$u(a)=-a^{2}$
$u(4)=-16$
$R_{4}\leq-\frac{1}{2}\lim\limits_{a \to \infty}\int_{-16}^{-a^{2}}e^{u}du$
$R_{4}\leq-\frac{1}{2}\lim\limits_{a \to \infty}[e^{u}]_{-16}^{-a^{2}}$
$R_{4}\leq-\frac{1}{2}\lim\limits_{a \to \infty}[e^{-a^{2}}-e^{-16}]=-\frac{1}{2}(\frac{1}{e^{\infty}}-\frac{1}{e^{16}})=-\frac{1}{2}(-\frac{1}{e^{16}})=\frac{1}{2e^{16}}$
$S\leq0.4048813985+\frac{1}{2e^{16}}\approx0.408814548$
