What is the sum from n=1 to infinity of (sin(400))^n ?

The sum from n=1 to infinity of (sin(400))^n is 1/(1-sin(400))-1

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sum from n=1 to infinity of (sin(400))^n

n = 1 \sum \infty (sin (400))^{n}

\frac{1}{1 - sin ( 400 )} - 1

Solution steps

n = 1 \sum \infty (sin (400))^{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 (400))^{n} - (sin (400))^{0}

n = 0 \sum \infty (sin (400))^{n} = \frac{1}{1 - sin ( 400 )}

= \frac{1}{1 - sin ( 400 )} - (sin (400))^{0}

Simplify

= \frac{1}{1 - sin ( 400 )} - 1

n = 0 \sum \infty \frac{n !}{n + 1}

n = 1 \sum \infty \frac{1}{7 n ^{3} + 3}

n = 1 \sum \infty 3 (\frac{1}{5})^{n}

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

i = 1 \sum \infty 3 e^{i}

What is the sum from n=1 to infinity of (sin(400))^n ?
The sum from n=1 to infinity of (sin(400))^n is 1/(1-sin(400))-1