What is the sum from n=1 to infinity of sin^n(1) ?

The sum from n=1 to infinity of sin^n(1) is 1/(1-sin(1))-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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)

Simplify

= \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)}

What is the sum from n=1 to infinity of sin^n(1) ?
The sum from n=1 to infinity of sin^n(1) is 1/(1-sin(1))-1