What is the sum from n=0 to infinity of n(5/6)^{n-1} ?

The sum from n=0 to infinity of n(5/6)^{n-1} is converges

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sum from n=0 to infinity of n(5/6)^{n-1}

n = 0 \sum \infty n (\frac{5}{6})^{n - 1}

converges

Solution steps

n = 0 \sum \infty n (\frac{5}{6})^{n - 1}

Simplify

(\frac{5}{6})^{n - 1} : (\frac{5}{6})^{n} (\frac{5}{6})^{- 1}

= n = 0 \sum \infty n (\frac{5}{6})^{n} (\frac{5}{6})^{- 1}

Apply the constant multiplication rule:

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

= (\frac{5}{6})^{- 1} \cdot n = 0 \sum \infty n (\frac{5}{6})^{n}

(\frac{5}{6})^{- 1} = \frac{6}{5}

= \frac{6}{5} \cdot n = 0 \sum \infty n (\frac{5}{6})^{n}

Apply Series Ratio Test:

converges

= \frac{6}{5} converges

= converges

\frac{d}{d x} (x^{2} y + 3 x \cdot y^{3} + x)

\int cot (6 x) d x

\frac{\partial}{\partial z} (x^{3})

x^{2} sin^{2} (5 x)

y = 5 e^{x}

What is the sum from n=0 to infinity of n(5/6)^{n-1} ?
The sum from n=0 to infinity of n(5/6)^{n-1} is converges