What is the sum from n=1 to infinity of (-7/9)^n ?

The sum from n=1 to infinity of (-7/9)^n is converges

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sum from n=1 to infinity of (-7/9)^n

n = 1 \sum \infty (- \frac{7}{9})^{n}

converges

Solution steps

n = 1 \sum \infty (- \frac{7}{9})^{n}

Apply Series Ratio Test:

converges

= converges

t^{2} \frac{d x}{d t} = \frac{1}{x ^{2}}, x (4) = 9

\int sin^{2} (u) d u

\frac{\partial}{\partial z} (x \cdot y^{2} \cdot e^{z})

\frac{d}{d x} (3 x^{2} + 6 x + cos (x))

f (x) = \frac{7}{x}

What is the sum from n=1 to infinity of (-7/9)^n ?
The sum from n=1 to infinity of (-7/9)^n is converges