What is the sum from n=1 to infinity of (-0.6)^n ?

The sum from n=1 to infinity of (-0.6)^n is converges

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sum from n=1 to infinity of (-0.6)^n

n = 1 \sum \infty (- 0.6)^{n}

converges

Solution steps

n = 1 \sum \infty (- 0.6)^{n}

Apply Series Ratio Test:

converges

= converges

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

k = 6 \sum \infty \frac{1}{( k - 5 ) ^{3}}

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

n = 1 \sum \infty \frac{1}{n ^{2} + 7 n + 1}

k = 0 \sum \infty (- \frac{1}{2})^{k}

What is the sum from n=1 to infinity of (-0.6)^n ?
The sum from n=1 to infinity of (-0.6)^n is converges