What is the sum from n=1 to infinity of (-0.45)^n ?

The sum from n=1 to infinity of (-0.45)^n is converges

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sum from n=1 to infinity of (-0.45)^n

n = 1 \sum \infty (- 0.45)^{n}

converges

Solution steps

n = 1 \sum \infty (- 0.45)^{n}

Apply Series Ratio Test:

converges

= converges

k = 1 \sum \infty \frac{k}{k + 1}

n = 0 \sum \infty (\frac{3}{π})^{n}

n = 1 \sum \infty (\frac{1}{9})^{n - 1}

n = 0 \sum \infty 8 \cdot (\frac{5}{8})^{n}

n = 0 \sum \infty 5 + \frac{15}{5} + \frac{45}{16}

What is the sum from n=1 to infinity of (-0.45)^n ?
The sum from n=1 to infinity of (-0.45)^n is converges