What is the sum from n=1 to infinity of (cos(2.8))^n ?

The sum from n=1 to infinity of (cos(2.8))^n is converges

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sum from n=1 to infinity of (cos(2.8))^n

n = 1 \sum \infty (cos (2.8))^{n}

converges

Solution steps

n = 1 \sum \infty (cos (2.8))^{n}

Simplify

cos (2.8) : - 0.94222 \dots

= n = 1 \sum \infty (- 0.94222 \dots)^{n}

Apply Series Ratio Test:

converges

= converges

\int_{- 3}^{3} 1 + x^{2} d x

\frac{d y}{d x}, y - 9 x^{3} + 8 = 0, (2, 8)

4^{x^{2}}

\int (csc (2 x)) d x

y^{^{'}} - 2 t y = 12 t^{2} e^{t^{2}}, y (0) = 5

What is the sum from n=1 to infinity of (cos(2.8))^n ?
The sum from n=1 to infinity of (cos(2.8))^n is converges