What is the sum from n=1 to infinity of (sin(100))^n ?

The sum from n=1 to infinity of (sin(100))^n is 1/(1-sin(100))-1

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sum from n=1 to infinity of (sin(100))^n

n = 1 \sum \infty (sin (100))^{n}

\frac{1}{1 - sin ( 100 )} - 1

Solution steps

n = 1 \sum \infty (sin (100))^{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 (100))^{n} - (sin (100))^{0}

n = 0 \sum \infty (sin (100))^{n} = \frac{1}{1 - sin ( 100 )}

= \frac{1}{1 - sin ( 100 )} - (sin (100))^{0}

Simplify

= \frac{1}{1 - sin ( 100 )} - 1

n = 1 \sum \infty (\frac{3}{10})^{n}

n = 1 \sum \infty \frac{n ^{2}}{2 n}

k = 1 \sum \infty \frac{k ^{k}}{k !}

k = 5 \sum \infty \frac{1}{k ^{2} - 5 k + 6}

n = 1 \sum \infty \frac{1}{n ^{\frac{1}{4}}}

What is the sum from n=1 to infinity of (sin(100))^n ?
The sum from n=1 to infinity of (sin(100))^n is 1/(1-sin(100))-1