What is the sum from k=2 to infinity of 3/(9^{k-1)} ?

The sum from k=2 to infinity of 3/(9^{k-1)} is converges

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sum from k=2 to infinity of 3/(9^{k-1)}

k = 2 \sum \infty \frac{3}{9 ^{k - 1}}

converges

Solution steps

k = 2 \sum \infty \frac{3}{9 ^{k - 1}}

Apply the constant multiplication rule:

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

= 3 \cdot k = 2 \sum \infty \frac{1}{9 ^{k - 1}}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\frac{d y}{d x} = \frac{y}{x ^{2}}

\frac{d}{d x} (\frac{1}{1 - e ^{- x}})

\int e cos (x) d x

\frac{d}{d y} (\frac{a x + b y}{c x + d y})

\frac{d}{d x} (\frac{2 x ^{\frac{3}{2}}}{3})

What is the sum from k=2 to infinity of 3/(9^{k-1)} ?
The sum from k=2 to infinity of 3/(9^{k-1)} is converges