What is the sum from n=1 to infinity of 8/(5^{n-1)} ?

The sum from n=1 to infinity of 8/(5^{n-1)} is converges

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sum from n=1 to infinity of 8/(5^{n-1)}

n = 1 \sum \infty \frac{8}{5 ^{n - 1}}

converges

Solution steps

n = 1 \sum \infty \frac{8}{5 ^{n - 1}}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{1}{5 ^{n - 1}}

Apply Series Ratio Test:

converges

= 8 converges

= converges

\frac{d}{d θ} (r (θ)), r (θ) = cos (θ) cot (θ)

f (x) = (x^{3} + 2) (x^{5} + 7 x^{2} + 3)

\frac{d}{d x} (\frac{x + 2}{x ^{2} + 5 x + 6})

f (x) = 2 + 5 x

(x - 3)^{2} \cdot (\frac{1}{2}) + 1

What is the sum from n=1 to infinity of 8/(5^{n-1)} ?
The sum from n=1 to infinity of 8/(5^{n-1)} is converges