sum from n=1 to infinity of (sin(3))^n

解

n = 1 \sum \infty (sin (3))^{n}

\frac{1}{1 - sin ( 3 )} - 1

十進法表記

0.16430 \dots

解答ステップ

n = 1 \sum \infty (sin (3))^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (sin (3))^{n} - (sin (3))^{0}

n = 0 \sum \infty (sin (3))^{n} = \frac{1}{1 - sin ( 3 )}

= \frac{1}{1 - sin ( 3 )} - (sin (3))^{0}

簡素化

= \frac{1}{1 - sin ( 3 )} - 1