What is the sum from n=1 to infinity of 9((1-3)/4)^n ?

The sum from n=1 to infinity of 9((1-3)/4)^n is converges

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sum from n=1 to infinity of 9((1-3)/4)^n

n = 1 \sum \infty 9 (\frac{1 - 3}{4})^{n}

converges

Solution steps

n = 1 \sum \infty 9 (\frac{1 - 3}{4})^{n}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty (\frac{1 - 3}{4})^{n}

\frac{1 - 3}{4} = - \frac{1}{2}

= 9 \cdot n = 1 \sum \infty (- \frac{1}{2})^{n}

Apply Series Ratio Test:

converges

= 9 converges

= converges

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What is the sum from n=1 to infinity of 9((1-3)/4)^n ?
The sum from n=1 to infinity of 9((1-3)/4)^n is converges