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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{\partial}{\partial x} (\frac{- y - x y}{( 1 + x ) ^{3}})

\int \frac{1}{x ^{2} + 25} d x

\int x (1 + x)^{4} d x

y^{^{'}} = t - y t

\frac{d y}{d x}, y = 2 x + 1

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