What is the sum from n=1 to infinity of 6e^{-8n} ?

The sum from n=1 to infinity of 6e^{-8n} is converges

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sum from n=1 to infinity of 6e^{-8n}

n = 1 \sum \infty 6 e^{- 8 n}

converges

Solution steps

n = 1 \sum \infty 6 e^{- 8 n}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty e^{- 8 n}

Apply Series Ratio Test:

converges

= 6 converges

= converges

x \to 0 lim (2 - x)

\int - e^{- s t} d t

\int \frac{2}{e ^{t}} d t

\frac{\partial}{\partial y} (\frac{1}{x ^{2} + y ^{2} - 1})

\frac{d y}{d x} = \frac{2 x y}{( x ^{2} - y ^{2} )}

What is the sum from n=1 to infinity of 6e^{-8n} ?
The sum from n=1 to infinity of 6e^{-8n} is converges