What is the sum from n=0 to infinity of 2/(3^{n-1)} ?

The sum from n=0 to infinity of 2/(3^{n-1)} is converges

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sum from n=0 to infinity of 2/(3^{n-1)}

n = 0 \sum \infty \frac{2}{3 ^{n - 1}}

converges

Solution steps

n = 0 \sum \infty \frac{2}{3 ^{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}{3 ^{n - 1}}

Apply Series Ratio Test:

converges

= 2 converges

= converges

\int - 3 x^{2} d x

\frac{\partial}{\partial z} (\frac{x ^{2}}{y} - x z)

\int \frac{x}{4 + x ^{4}} d x

\frac{\partial}{\partial y} (\frac{7}{x ^{2} + y ^{2}})

\frac{( x ^{2} + 32 x )}{8}, at x = 4

What is the sum from n=0 to infinity of 2/(3^{n-1)} ?
The sum from n=0 to infinity of 2/(3^{n-1)} is converges