What is the sum from n=1 to infinity of (sin(2))^n ?

The sum from n=1 to infinity of (sin(2))^n is converges

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sum from n=1 to infinity of (sin(2))^n

n = 1 \sum \infty (sin (2))^{n}

converges

Solution steps

n = 1 \sum \infty (sin (2))^{n}

Simplify

= n = 1 \sum \infty sin^{n} (2)

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{5 ^{n} - n}

i = 1 \sum \infty 5 \cdot 0. 2^{i - 1}

n = 1 \sum \infty \frac{3}{5 ^{n + 2}}

n = 0 \sum \infty ln (1 - n^{2})

n = 0 \sum \infty 5^{- 2 n + 1} 3^{n}

What is the sum from n=1 to infinity of (sin(2))^n ?
The sum from n=1 to infinity of (sin(2))^n is converges