What is the sum from n=2 to infinity of (-0.25)^n ?

The sum from n=2 to infinity of (-0.25)^n is converges

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sum from n=2 to infinity of (-0.25)^n

n = 2 \sum \infty (- 0.25)^{n}

converges

Solution steps

n = 2 \sum \infty (- 0.25)^{n}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{n}{4 n + 1}

n = 1 \sum \infty sin (\frac{nπ}{2})

n = 0 \sum \infty (\frac{4}{5})^{2 n}

n = 1 \sum \infty 5 ln (\frac{n}{n + 1})

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

What is the sum from n=2 to infinity of (-0.25)^n ?
The sum from n=2 to infinity of (-0.25)^n is converges