What is the sum from n=2 to infinity of 6e^{-3n} ?

The sum from n=2 to infinity of 6e^{-3n} is converges

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sum from n=2 to infinity of 6e^{-3n}

n = 2 \sum \infty 6 e^{- 3 n}

converges

Solution steps

n = 2 \sum \infty 6 e^{- 3 n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 2 \sum \infty e^{- 3 n}

Apply Series Ratio Test:

converges

= 6 converges

= converges

n = 0 \sum \infty n (e^{- n})

n = 1 \sum \infty \frac{n !}{10 9 ^{n}}

n = 2 \sum \infty \frac{1}{n ^{2} + 4}

n = 1 \sum \infty \frac{2022}{9 ^{n}}

n = 1 \sum \infty n^{- 4}

What is the sum from n=2 to infinity of 6e^{-3n} ?
The sum from n=2 to infinity of 6e^{-3n} is converges