What is the sum from n=-4 to infinity of (-4/5)^n ?

The sum from n=-4 to infinity of (-4/5)^n is converges

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sum from n=-4 to infinity of (-4/5)^n

n = - 4 \sum \infty (- \frac{4}{5})^{n}

converges

Solution steps

n = - 4 \sum \infty (- \frac{4}{5})^{n}

Apply Series Ratio Test:

converges

= converges

y = \frac{1}{3} x^{3} - 2 x + 2, (- 3, - 1)

\int \frac{x - 3}{x ^{2} - 6 x + 10} d x

\frac{d}{d x} (\frac{x y}{x - y})

\frac{d}{d x} (3 x - \frac{5}{x})

y = \frac{1}{2} x^{2}, 2 x

What is the sum from n=-4 to infinity of (-4/5)^n ?
The sum from n=-4 to infinity of (-4/5)^n is converges