What is the sum from k=2 to infinity of (-0.75)^k ?

The sum from k=2 to infinity of (-0.75)^k is converges

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sum from k=2 to infinity of (-0.75)^k

k = 2 \sum \infty (- 0.75)^{k}

converges

Solution steps

k = 2 \sum \infty (- 0.75)^{k}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty n^{- 0.3}

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

x = 0 \sum \infty \frac{x}{2 ^{x}}

x = 0 \sum \infty \frac{1}{x !}

n = 4 \sum \infty \frac{2}{n ( n - 1 )}

What is the sum from k=2 to infinity of (-0.75)^k ?
The sum from k=2 to infinity of (-0.75)^k is converges