What is the sum from k=2 to infinity of (-0.95)^k ?

The sum from k=2 to infinity of (-0.95)^k is converges

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sum from k=2 to infinity of (-0.95)^k

k = 2 \sum \infty (- 0.95)^{k}

converges

Solution steps

k = 2 \sum \infty (- 0.95)^{k}

Apply Series Ratio Test:

converges

= converges

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

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

n = 1 \sum \infty n - 1

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

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

What is the sum from k=2 to infinity of (-0.95)^k ?
The sum from k=2 to infinity of (-0.95)^k is converges