What is the sum from n=1 to infinity of (-0.7)^{n-1} ?

The sum from n=1 to infinity of (-0.7)^{n-1} is converges

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sum from n=1 to infinity of (-0.7)^{n-1}

n = 1 \sum \infty (- 0.7)^{n - 1}

converges

Solution steps

n = 1 \sum \infty (- 0.7)^{n - 1}

Apply Series Ratio Test:

converges

= converges

\frac{\partial}{\partial t} (\frac{x}{t})

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x \to 1 lim (e^{6 x^{5} - 6 x})

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y = f (u) = u (y, x)^{3} - 4 y u (y, x) = g (x) = x^{3} + 3

What is the sum from n=1 to infinity of (-0.7)^{n-1} ?
The sum from n=1 to infinity of (-0.7)^{n-1} is converges