What is the sum from n=2 to infinity of (-0.45)^n ?

The sum from n=2 to infinity of (-0.45)^n is converges

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sum from n=2 to infinity of (-0.45)^n

n = 2 \sum \infty (- 0.45)^{n}

converges

Solution steps

n = 2 \sum \infty (- 0.45)^{n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty 7 - 4 (0.8)^{n}

n = 0 \sum \infty \frac{1}{9 - n ^{2}}

n = 1 \sum \infty 3 (\frac{1}{7})^{n + 1}

k = 3 \sum \infty \frac{k - 1}{2 k + 1}

n = 1 \sum \infty 5 (\frac{3}{2})^{n}

What is the sum from n=2 to infinity of (-0.45)^n ?
The sum from n=2 to infinity of (-0.45)^n is converges