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

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