What is the sum from n=2 to infinity of (-0.5)^n ?

The sum from n=2 to infinity of (-0.5)^n is converges

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sum from n=2 to infinity of (-0.5)^n

n = 2 \sum \infty (- 0.5)^{n}

converges

Solution steps

n = 2 \sum \infty (- 0.5)^{n}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{1}{4 + n}

n = 1 \sum \infty (\frac{1}{3})^{n + 1}

n = 0 \sum \infty \frac{4}{n + 3}

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

n = 2 \sum \infty (1 + \frac{1}{3})^{- n}

What is the sum from n=2 to infinity of (-0.5)^n ?
The sum from n=2 to infinity of (-0.5)^n is converges