Answer
$error \leq 6.4 \times 10^{-8}$
and
first ten terms $\approx 0.0739293$
Work Step by Step
$\Sigma_{n=1}^{\infty} 5^{-n}cos^{2}n=\Sigma_{n=1}^{\infty}\frac{cos^{2}n}{5^{n}}$
$T_{n}=\int_{10}^{\infty}\frac{1}{5^{x}}dx=\lim\limits_{t \to \infty} \int_{10}^{t}\frac{1}{5^{x}}dx$
$=\lim\limits_{t \to \infty} [-\frac{5^{-x}}{ln5}]_{10}^{t}$
$\approx 6.4 \times 10^{-8}$
Using a calculator or adding 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 $\approx 0.0739293$
