What is the sum from k=2 to infinity of (-0.35)^k ?

The sum from k=2 to infinity of (-0.35)^k is converges

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sum from k=2 to infinity of (-0.35)^k

k = 2 \sum \infty (- 0.35)^{k}

converges

Solution steps

k = 2 \sum \infty (- 0.35)^{k}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{7 ^{n}}{1 0 ^{n}}

k = 5 \sum \infty 2 (- \frac{1}{4})^{k}

n = 4 \sum \infty 4 (6 - 2 n)

n = 1 \sum \infty 3^{n} 4^{- n + 1}

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

What is the sum from k=2 to infinity of (-0.35)^k ?
The sum from k=2 to infinity of (-0.35)^k is converges