What is the sum from n=1 to infinity of 2e^{-9n} ?

The sum from n=1 to infinity of 2e^{-9n} is converges

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sum from n=1 to infinity of 2e^{-9n}

n = 1 \sum \infty 2 e^{- 9 n}

converges

Solution steps

n = 1 \sum \infty 2 e^{- 9 n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 1 \sum \infty e^{- 9 n}

Apply Series Ratio Test:

converges

= 2 converges

= converges

\int t e^{8 t} d t

f (x) = 5 x^{- \frac{3}{5}}

y = x^{3} - 7 x^{2} + 8 x + 1

\int \frac{64}{x ^{3} - 8 x ^{2}} d x

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

What is the sum from n=1 to infinity of 2e^{-9n} ?
The sum from n=1 to infinity of 2e^{-9n} is converges