Answer
$n\geq16$
Work Step by Step
$\Sigma_{n=1}^{\infty}e^{-\frac{n}{2}}$
The function $e^{-\frac{n}{2}}$ is positive, continuous, and ultimately decreasing. Therefore we can use the integral test to find the minimum value of $n$ for $R_{n}\leq0.001$:
$R_{n}\leq\int_{n}^{\infty}e^{-\frac{n}{2}}dn$
$R_{n}\leq\lim\limits_{a \to \infty}\int_{n}^{a}e^{-\frac{n}{2}}dn$
$let$ $u=-\frac{n}{2}$
$du=-\frac{1}{2}dn$
$-2du=dn$
$u(n)=-\frac{n}{2}$
$u(a)=-\frac{a}{2}$
$R_{n}\leq-2\lim\limits_{a \to \infty}\int_{-\frac{n}{2}}^{-\frac{a}{2}}e^{u}du$
$R_{n}\leq-2\lim\limits_{a \to \infty}[e^{u}]_{-\frac{n}{2}}^{-\frac{a}{2}}$
$R_{n}\leq-2\lim\limits_{a \to \infty}[e^{-\frac{a}{2}}-e^{-\frac{n}{2}}]=-2(e^{-\frac{\infty}{2}}-e^{-\frac{n}{2}})=-2(\frac{1}{e^{\infty}}-\frac{1}{e^{\frac{n}{2}}})=\frac{2}{e^{\frac{n}{2}}}$
$\frac{2}{e^{\frac{n}{2}}}\leq\frac{1}{1000}$
$2000\leq e^{\frac{n}{2}}$
$ln2000\leq \frac{n}{2}lne$
$ln2000\leq \frac{n}{2}$
$2ln2000\leq n$
Round up to the nearest integer:
$n\geq16$
