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

Lösung

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

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

Dezimale

0.16430 \dots

Schritte zur Lösung

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

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} (3) - sin^{0} (3)

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

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

Vereinfache

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

n = 1 \sum \infty sin^{n} (9)

n = 1 \sum \infty \frac{5}{n ^{n}}

n = 1 \sum \infty \frac{ln ( 4 n )}{n}

n = 1 \sum \infty \frac{3 ^{n}}{3 + n !}

n = 1 \sum \infty (\frac{4}{5})^{n - 1}