What is the sum from n=1 to infinity of 9/((-2)^n) ?

The sum from n=1 to infinity of 9/((-2)^n) is converges

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sum from n=1 to infinity of 9/((-2)^n)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 9 converges

= converges

x \to 1 - lim (e^{\frac{1}{x - 1}})

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What is the sum from n=1 to infinity of 9/((-2)^n) ?
The sum from n=1 to infinity of 9/((-2)^n) is converges