What is the sum from n=1 to infinity of (cos(28))^n ?

The sum from n=1 to infinity of (cos(28))^n is converges

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sum from n=1 to infinity of (cos(28))^n

n = 1 \sum \infty (cos (28))^{n}

converges

Solution steps

n = 1 \sum \infty (cos (28))^{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 (cos (28))^{n} - (cos (28))^{0}

n = 0 \sum \infty (cos (28))^{n} =

converges

= converges - (cos (28))^{0}

= converges

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What is the sum from n=1 to infinity of (cos(28))^n ?
The sum from n=1 to infinity of (cos(28))^n is converges