What is the sum from n=0 to infinity of 2/(5^{n-1)} ?

The sum from n=0 to infinity of 2/(5^{n-1)} is converges

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sum from n=0 to infinity of 2/(5^{n-1)}

n = 0 \sum \infty \frac{2}{5 ^{n - 1}}

converges

Solution steps

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

y = sin (sin (x)), (3 π, 0)

(- x)^{^{'}}

\frac{d}{d x} ((\frac{4}{5} x)^{3})

\frac{\partial}{\partial y} (x^{2} cos (4 y))

\frac{x ^{2}}{7}

What is the sum from n=0 to infinity of 2/(5^{n-1)} ?
The sum from n=0 to infinity of 2/(5^{n-1)} is converges