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

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

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sum from n=1 to infinity of (-3)^{-n}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{\partial}{\partial x} (4 y)

\frac{\partial}{\partial x} (6 x + 5 y)

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

n = 1 \sum \infty 3 e^{- 9 n}

f (x) = \frac{1}{x ^{2}}, [2, 6]

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