What is the sum from n=0 to infinity of 2^{-2-2n} ?

The sum from n=0 to infinity of 2^{-2-2n} is converges

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sum from n=0 to infinity of 2^{-2-2n}

n = 0 \sum \infty 2^{- 2 - 2 n}

converges

Solution steps

n = 0 \sum \infty 2^{- 2 - 2 n}

Apply exponent rule:

a^{- n} = \frac{1}{a ^{n}}

2^{- 2 - 2 n} = \frac{1}{2 ^{2 + 2 n}}

= n = 0 \sum \infty \frac{1}{2 ^{2 + 2 n}}

Apply Series Ratio Test:

converges

= converges

π \cdot c \cdot \frac{cos ( \frac{π x}{L} )}{L}

\int 6 x^{2} sin (\frac{1}{2} x) d x

\frac{x ^{2} - 1}{x + 3}

e^{2 - t}

(t - 4)^{2}

What is the sum from n=0 to infinity of 2^{-2-2n} ?
The sum from n=0 to infinity of 2^{-2-2n} is converges