What is the sum from n=2 to infinity of (-0.1)^n ?

The sum from n=2 to infinity of (-0.1)^n is converges

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sum from n=2 to infinity of (-0.1)^n

n = 2 \sum \infty (- 0.1)^{n}

converges

Solution steps

n = 2 \sum \infty (- 0.1)^{n}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty (\frac{4}{3})^{- n}

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

n = 0 \sum \infty \frac{- 1}{n}

n = 1 \sum \infty \frac{6}{n ^{2} + 3}

n = 1 \sum \infty \frac{1}{n ^{\frac{9}{2}}}

What is the sum from n=2 to infinity of (-0.1)^n ?
The sum from n=2 to infinity of (-0.1)^n is converges