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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

y^{^{''}} + 4 y^{^{'}} + 4 y = 2 x + 6

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x} sin (x)

\int 0.06 \cdot e^{\frac{t}{300}} d t

\frac{d}{d x} (ln (\frac{2 x + 1}{x - 1}))

n = 1 \sum \infty \frac{7}{( n + 1 ) !}

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