## Calculus 8th Edition

$error \leq 6.4 \times 10^{-8}$ and first ten terms $\approx 0.0739293$
$\Sigma_{n=1}^{\infty} 5^{-n}cos^{2}n=\Sigma_{n=1}^{\infty}\frac{cos^{2}n}{5^{n}}$ $T_{n}=\int_{10}^{\infty}\frac{1}{5^{x}}dx=\lim\limits_{t \to \infty} \int_{10}^{t}\frac{1}{5^{x}}dx$ $=\lim\limits_{t \to \infty} [-\frac{5^{-x}}{ln5}]_{10}^{t}$ $\approx 6.4 \times 10^{-8}$ Using a calculator or by hand we can sum the first 10 terms: $\Sigma_{n=1}^{\infty}\frac{cos^{2}n}{5^{n}}\approx 0.0739293$ Hence, $error \leq 6.4 \times 10^{-8}$ and first ten terms $\approx 0.0739293$