What is the sum from n=1 to infinity of ((-1)/3)^n ?

The sum from n=1 to infinity of ((-1)/3)^n is converges

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sum from n=1 to infinity of ((-1)/3)^n

n = 1 \sum \infty (\frac{- 1}{3})^{n}

converges

Solution steps

n = 1 \sum \infty (\frac{- 1}{3})^{n}

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty \frac{4}{n ^{2} + n ^{3}}

k = 3 \sum \infty \frac{5}{k} - \frac{5}{k + 1}

k = 0 \sum \infty (0.9)^{k}

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

What is the sum from n=1 to infinity of ((-1)/3)^n ?
The sum from n=1 to infinity of ((-1)/3)^n is converges