What is the sum from n=1 to infinity of e^{-6n} ?

The sum from n=1 to infinity of e^{-6n} is converges

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sum from n=1 to infinity of e^{-6n}

n = 1 \sum \infty e^{- 6 n}

converges

Solution steps

n = 1 \sum \infty e^{- 6 n}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (2 x + \frac{x ^{2} - x}{x ^{3} + x + 1})

sin (4 x) cos (2 x)

\int \frac{1.2 x}{x ^{2} x ^{4} - 9} d x

(\frac{sin ( x )}{1 + x})^{^{'}}

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

What is the sum from n=1 to infinity of e^{-6n} ?
The sum from n=1 to infinity of e^{-6n} is converges