What is the sum from n=1 to infinity of e^{-8n} ?

The sum from n=1 to infinity of e^{-8n} is converges

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sum from n=1 to infinity of e^{-8n}

n = 1 \sum \infty e^{- 8 n}

converges

Solution steps

n = 1 \sum \infty e^{- 8 n}

Apply Series Ratio Test:

converges

= converges

t \to π lim (cos (\frac{2 t}{3}))

\frac{d}{d x} (\frac{3}{x ^{2}} + 10 x^{3} - 2 sin (3 x))

\int \frac{8}{7 + 8 x} d x

f (x) = - x^{2} + 4, y + 2 x - 4 = 0

7 - 3 x

What is the sum from n=1 to infinity of e^{-8n} ?
The sum from n=1 to infinity of e^{-8n} is converges