What is the sum from n=0 to infinity of (-7/9)^n ?

The sum from n=0 to infinity of (-7/9)^n is 9/16

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sum from n=0 to infinity of (-7/9)^n

n = 0 \sum \infty (- \frac{7}{9})^{n}

\frac{9}{16}

Solution steps

n = 0 \sum \infty (- \frac{7}{9})^{n}

Apply Series Geometric Test:

\frac{9}{16}

= \frac{9}{16}

n = 0 \sum \infty ln (1 + n^{2})

n = 0 \sum \infty \frac{1}{3 ^{n} + n}

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

k = 1 \sum \infty k^{2}

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

What is the sum from n=0 to infinity of (-7/9)^n ?
The sum from n=0 to infinity of (-7/9)^n is 9/16