What is the sum from n=0 to infinity of ((-2)/3)^n ?

The sum from n=0 to infinity of ((-2)/3)^n is 3/5

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sum from n=0 to infinity of ((-2)/3)^n

n = 0 \sum \infty (\frac{- 2}{3})^{n}

\frac{3}{5}

Solution steps

n = 0 \sum \infty (\frac{- 2}{3})^{n}

Apply Series Geometric Test:

\frac{3}{5}

= \frac{3}{5}

n = 7 \sum \infty 2^{- n}

n = 0 \sum \infty (\frac{1}{2})^{n - 1}

n = 0 \sum \infty sin (\frac{1}{n ^{1}})

n = 0 \sum \infty (\frac{11}{12})^{n}

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

What is the sum from n=0 to infinity of ((-2)/3)^n ?
The sum from n=0 to infinity of ((-2)/3)^n is 3/5