What is the sum from n=2 to infinity of (-0.7)^n ?

The sum from n=2 to infinity of (-0.7)^n is converges

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sum from n=2 to infinity of (-0.7)^n

n = 2 \sum \infty (- 0.7)^{n}

converges

Solution steps

n = 2 \sum \infty (- 0.7)^{n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1 - n}{n 2 ^{n}}

n = 1 \sum \infty \frac{1}{6 + e ^{- n}}

n = 2 \sum \infty \frac{2}{n ln ( n )}

n = 1 \sum \infty 5^{- 3 n}

n = 1 \sum \infty \frac{1}{e ^{4 n}}

What is the sum from n=2 to infinity of (-0.7)^n ?
The sum from n=2 to infinity of (-0.7)^n is converges