What is the sum from n=1 to infinity of 2/(5^{n-1)} ?

The sum from n=1 to infinity of 2/(5^{n-1)} is converges

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sum from n=1 to infinity of 2/(5^{n-1)}

n = 1 \sum \infty \frac{2}{5 ^{n - 1}}

converges

Solution steps

n = 1 \sum \infty \frac{2}{5 ^{n - 1}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{5 ^{n - 1}}

Apply Series Ratio Test:

converges

= 2 converges

= converges

\int 2 x e^{x^{2} - 1} d x

\frac{d}{d x} ((x^{2} + 2)^{\frac{1}{2}})

\int (x + \frac{1}{2 x}) d x

\int - cos (x) (2 x + 1) d x

\frac{d}{d x} (\frac{ln ( x )}{x ^{9}})

What is the sum from n=1 to infinity of 2/(5^{n-1)} ?
The sum from n=1 to infinity of 2/(5^{n-1)} is converges