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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 9 converges

= converges

n = 0 \sum \infty e^{2 l n (n)}

k = 1 \sum \infty \frac{6}{k} - \frac{6}{k + 1}

n = 0 \sum \infty 5 (\frac{3}{π})^{n}

n = 1 \sum \infty (\frac{- 1}{7})^{n}

n = 0 \sum \infty (\frac{1}{2})^{n - 2}

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