What is the sum from n=2 to infinity of (-0.6)^n ?

The sum from n=2 to infinity of (-0.6)^n is converges

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sum from n=2 to infinity of (-0.6)^n

n = 2 \sum \infty (- 0.6)^{n}

converges

Solution steps

n = 2 \sum \infty (- 0.6)^{n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty cos (\frac{14}{n})

n = 1 \sum \infty \frac{n}{( n + 1 ) 2 ^{n}}

n = 1 \sum \infty \frac{1}{n ^{2} + 10}

n = 1 \sum \infty \frac{8}{n ^{2} - 1}

k = 1 \sum \infty \frac{3}{4 ^{k}}

What is the sum from n=2 to infinity of (-0.6)^n ?
The sum from n=2 to infinity of (-0.6)^n is converges