What is the sum from n=0 to infinity of 8e^{-n} ?

The sum from n=0 to infinity of 8e^{-n} is 8*e/(e-1)

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sum from n=0 to infinity of 8e^{-n}

n = 0 \sum \infty 8 e^{- n}

8 \cdot \frac{e}{e - 1}

Solution steps

n = 0 \sum \infty 8 e^{- n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 0 \sum \infty e^{- n}

Simplify

e^{- n} : \frac{1}{e ^{n}}

= 8 \cdot n = 0 \sum \infty \frac{1}{e ^{n}}

Apply Series Geometric Test:

\frac{e}{e - 1}

= 8 \cdot \frac{e}{e - 1}

n = 2 \sum \infty (- 0.1)^{n}

n = 0 \sum \infty (\frac{4}{3})^{- n}

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

n = 0 \sum \infty \frac{- 1}{n}

n = 1 \sum \infty \frac{6}{n ^{2} + 3}

What is the sum from n=0 to infinity of 8e^{-n} ?
The sum from n=0 to infinity of 8e^{-n} is 8*e/(e-1)