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

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

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

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

converges

Solution steps

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

Simplify

3^{n + 1} : 3^{n} \cdot 3

= n = 2 \sum \infty \frac{n}{3 ^{n} \cdot 3}

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= \frac{1}{3} converges

= converges

ln (\frac{8 - x}{8 + x})

72 x - \frac{4 x ^{3}}{3}

\frac{\partial}{\partial y} (\frac{y}{1 + x ^{2}})

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

\frac{d}{d θ} (2 cos (θ))

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