What is the sum from n=1 to infinity of-9(1/6)^{n-1} ?

The sum from n=1 to infinity of-9(1/6)^{n-1} is converges

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sum from n=1 to infinity of-9(1/6)^{n-1}

n = 1 \sum \infty - 9 (\frac{1}{6})^{n - 1}

converges

Solution steps

n = 1 \sum \infty - 9 (\frac{1}{6})^{n - 1}

Apply the constant multiplication rule:

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

= - 9 \cdot n = 1 \sum \infty (\frac{1}{6})^{n - 1}

Apply Series Ratio Test:

converges

= - 9 converges

= converges

What is the sum from n=1 to infinity of-9(1/6)^{n-1} ?
The sum from n=1 to infinity of-9(1/6)^{n-1} is converges