What is the sum from n=0 to infinity of 4/(5^{n-1)} ?

The sum from n=0 to infinity of 4/(5^{n-1)} is converges

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sum from n=0 to infinity of 4/(5^{n-1)}

n = 0 \sum \infty \frac{4}{5 ^{n - 1}}

converges

Solution steps

n = 0 \sum \infty \frac{4}{5 ^{n - 1}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{5 ^{n - 1}}

Apply Series Ratio Test:

converges

= 4 converges

= converges

x \to - 2 - lim (\frac{x}{x + 2})

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e^{- t} \cdot sin (t)

\int x d (ln (x)) d x

What is the sum from n=0 to infinity of 4/(5^{n-1)} ?
The sum from n=0 to infinity of 4/(5^{n-1)} is converges