What is the sum from n=1 to infinity of 4^{-2n} ?

The sum from n=1 to infinity of 4^{-2n} is converges

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sum from n=1 to infinity of 4^{-2n}

n = 1 \sum \infty 4^{- 2 n}

converges

Solution steps

n = 1 \sum \infty 4^{- 2 n}

Apply Series Ratio Test:

converges

= converges

y^{^{'}} = y \cdot (x + 1)

x + 2 sin (x)

f (x) = \frac{x}{x ^{2} + 9}

\int_{0}^{π} e^{x} cos (x) d x

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

What is the sum from n=1 to infinity of 4^{-2n} ?
The sum from n=1 to infinity of 4^{-2n} is converges