What is the sum from n=0 to infinity of (-9/10)^n ?

The sum from n=0 to infinity of (-9/10)^n is 10/19

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sum from n=0 to infinity of (-9/10)^n

n = 0 \sum \infty (- \frac{9}{10})^{n}

\frac{10}{19}

Solution steps

n = 0 \sum \infty (- \frac{9}{10})^{n}

Apply Series Geometric Test:

\frac{10}{19}

= \frac{10}{19}

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

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

n = 0 \sum \infty 5 (- \frac{2}{3})^{n}

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

n = 1 \sum \infty \frac{99}{10 0 ^{n}}

What is the sum from n=0 to infinity of (-9/10)^n ?
The sum from n=0 to infinity of (-9/10)^n is 10/19