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

解答

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

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

十进制

5.30799 \dots

求解步骤

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

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^{n} (1) - sin^{0} (1)

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

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

化简

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

n = 1 \sum \infty \frac{n ^{6}}{n !}

n = 1 \sum \infty (\frac{2}{4})^{n - 1}

n = 0 \sum \infty \frac{e ^{n}}{2 ^{n}}

n = 0 \sum \infty (\frac{1}{21})^{n}

n = 2 \sum \infty n^{l n (n)}