What is the sum from n=1 to infinity of (-0.2)^{n-1} ?

The sum from n=1 to infinity of (-0.2)^{n-1} is converges

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sum from n=1 to infinity of (-0.2)^{n-1}

n = 1 \sum \infty (- 0.2)^{n - 1}

converges

Solution steps

n = 1 \sum \infty (- 0.2)^{n - 1}

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (x^{2} 2 x - 7)

\int \frac{e ^{8 x}}{e ^{8 x} + 7} d x

\int \frac{1}{cos ^{3} ( x )} d x

x \to - \infty lim (x)

y = x^{11} e^{x}

What is the sum from n=1 to infinity of (-0.2)^{n-1} ?
The sum from n=1 to infinity of (-0.2)^{n-1} is converges