What is the sum from n=1 to infinity of 5e^{-6n} ?

The sum from n=1 to infinity of 5e^{-6n} is converges

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sum from n=1 to infinity of 5e^{-6n}

n = 1 \sum \infty 5 e^{- 6 n}

converges

Solution steps

n = 1 \sum \infty 5 e^{- 6 n}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty e^{- 6 n}

Apply Series Ratio Test:

converges

= 5 converges

= converges

\int_{0}^{1} 3 (1 + x)^{8} d x

\int csc^{4} (5 x) cot^{4} (5 x) d x

\frac{d}{d x} (2 x e^{- x^{2}})

(y^{16}) d y = \frac{( 1 + x )}{x} d x, y (1) = 5

\frac{( s ^{2} + s )}{2 s ^{2} + 2 s + 1}

What is the sum from n=1 to infinity of 5e^{-6n} ?
The sum from n=1 to infinity of 5e^{-6n} is converges