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

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

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

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

\frac{3}{5}

Solution steps

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

Apply Series Geometric Test:

\frac{3}{5}

= \frac{3}{5}

n = 0 \sum \infty \frac{3}{5 ^{n + 2}}

n = 1 \sum \infty \frac{e}{π ^{n}}

n = 1 \sum \infty ln (n^{2})

n = 1 \sum \infty n e^{2 n}

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

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