What is the sum from n=2 to infinity of (-0.8)^n ?

The sum from n=2 to infinity of (-0.8)^n is converges

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sum from n=2 to infinity of (-0.8)^n

n = 2 \sum \infty (- 0.8)^{n}

converges

Solution steps

n = 2 \sum \infty (- 0.8)^{n}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty 7 (\frac{3}{4})^{n}

n = 0 \sum \infty \frac{5}{n}

n = 0 \sum \infty ∣ n ∣

n = 0 \sum \infty (\frac{7}{5})^{n}

n = 0 \sum \infty \frac{sin ( \frac{1}{n} )}{n}

What is the sum from n=2 to infinity of (-0.8)^n ?
The sum from n=2 to infinity of (-0.8)^n is converges