What is the sum from n=0 to infinity of 8(1/9)^n ?

The sum from n=0 to infinity of 8(1/9)^n is 9

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sum from n=0 to infinity of 8(1/9)^n

n = 0 \sum \infty 8 (\frac{1}{9})^{n}

9

Solution steps

n = 0 \sum \infty 8 (\frac{1}{9})^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 8 \cdot n = 0 \sum \infty (\frac{1}{9})^{n}

Apply Series Geometric Test:

\frac{9}{8}

= 8 \cdot \frac{9}{8}

Simplify

8 \cdot \frac{9}{8} : 9

= 9

n = 0 \sum \infty e^{- 18 n}

k = 1 \sum \infty (\frac{2}{3})^{k}

n = 1 \sum \infty (- 0.01 + 1)^{n}

n = 0 \sum \infty \frac{n}{5 n - 3}

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

What is the sum from n=0 to infinity of 8(1/9)^n ?
The sum from n=0 to infinity of 8(1/9)^n is 9