What is the sum from n=1 to infinity of (cos(29))^n ?

The sum from n=1 to infinity of (cos(29))^n is converges

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sum from n=1 to infinity of (cos(29))^n

n = 1 \sum \infty (cos (29))^{n}

converges

Solution steps

n = 1 \sum \infty (cos (29))^{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 (29))^{n} - (cos (29))^{0}

n = 0 \sum \infty (cos (29))^{n} =

converges

= converges - (cos (29))^{0}

= converges

e^{\frac{2}{x}}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{3 + x})

f (x) = (x^{2} + 3)^{4}

\int \frac{1}{( 1 - x ) ^{3} x ^{2} - 2 x - 1} d x

What is the sum from n=1 to infinity of (cos(29))^n ?
The sum from n=1 to infinity of (cos(29))^n is converges