What is the sum from k=0 to infinity of (-4/5)^k ?

The sum from k=0 to infinity of (-4/5)^k is 5/9

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sum from k=0 to infinity of (-4/5)^k

k = 0 \sum \infty (- \frac{4}{5})^{k}

\frac{5}{9}

Solution steps

k = 0 \sum \infty (- \frac{4}{5})^{k}

Apply Series Geometric Test:

\frac{5}{9}

= \frac{5}{9}

n = 0 \sum \infty 7 (\frac{4}{5})^{n}

n = 1 \sum \infty \frac{n ^{2}}{8 ^{n}}

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

n = 0 \sum \infty (\frac{10}{11})^{n}

i = 0 \sum \infty \frac{1}{2 ^{i}}

What is the sum from k=0 to infinity of (-4/5)^k ?
The sum from k=0 to infinity of (-4/5)^k is 5/9