What is the sum from n=1 to infinity of 2e^{-8n} ?

The sum from n=1 to infinity of 2e^{-8n} is converges

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sum from n=1 to infinity of 2e^{-8n}

n = 1 \sum \infty 2 e^{- 8 n}

converges

Solution steps

n = 1 \sum \infty 2 e^{- 8 n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty e^{- 8 n}

Apply Series Ratio Test:

converges

= 2 converges

= converges

- \frac{6}{x ^{2}}

\frac{d}{d x} (x^{5} ln (8 x))

\int \frac{5 x ^{2}}{( 49 + x ^{2} ) ^{2}} d x

y = \frac{2 x}{x - 1}, (- 1, 1)

\frac{d}{d x} (x^{a x})

What is the sum from n=1 to infinity of 2e^{-8n} ?
The sum from n=1 to infinity of 2e^{-8n} is converges