What is the sum from n=1 to infinity of 2^{-3n} ?

The sum from n=1 to infinity of 2^{-3n} is converges

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sum from n=1 to infinity of 2^{-3n}

n = 1 \sum \infty 2^{- 3 n}

converges

Solution steps

n = 1 \sum \infty 2^{- 3 n}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d y} (9 x e^{- \frac{1}{y ^{3}}})

\int e^{x^{2} + 3} d x

y = 8 x - x^{2}, (1, 7)

y^{^{''}} - 2 y^{^{'}} - 8 y = 0 y (0) = - 2, y^{^{'}} (0) = 4

\int \frac{t + 2 t ^{2}}{t} d t

What is the sum from n=1 to infinity of 2^{-3n} ?
The sum from n=1 to infinity of 2^{-3n} is converges