What is the sum from n=0 to infinity of (sin(9))^n ?

The sum from n=0 to infinity of (sin(9))^n is 1/(1-sin(9))

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sum from n=0 to infinity of (sin(9))^n

n = 0 \sum \infty (sin (9))^{n}

\frac{1}{1 - sin ( 9 )}

Solution steps

n = 0 \sum \infty (sin (9))^{n}

Apply Series Geometric Test:

\frac{1}{1 - sin ( 9 )}

= \frac{1}{1 - sin ( 9 )}

n = 2 \sum \infty \frac{n}{( ln ( n ) ) n}

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

k = 1 \sum \infty 90 (0.1)^{k}

n = 0 \sum \infty \frac{4}{9 - n}

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

What is the sum from n=0 to infinity of (sin(9))^n ?
The sum from n=0 to infinity of (sin(9))^n is 1/(1-sin(9))