What is the sum from n=2 to infinity of 2e^{-4n} ?

The sum from n=2 to infinity of 2e^{-4n} is converges

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sum from n=2 to infinity of 2e^{-4n}

n = 2 \sum \infty 2 e^{- 4 n}

converges

Solution steps

n = 2 \sum \infty 2 e^{- 4 n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 2 \sum \infty e^{- 4 n}

Apply Series Ratio Test:

converges

= 2 converges

= converges

n = 1 \sum \infty \frac{i}{5 ^{n}}

n = 1 \sum \infty \frac{1}{cos ( \frac{1}{n} )}

n = 0 \sum \infty (n + 8)^{2}

n = 1 \sum \infty \frac{- 3 n}{2 n + 1}

n = 2 \sum \infty \frac{1}{n ^{5}}

What is the sum from n=2 to infinity of 2e^{-4n} ?
The sum from n=2 to infinity of 2e^{-4n} is converges