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

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

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

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

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

Solution steps

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

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

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

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

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

Simplify

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

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