What is the sum from n=0 to infinity of 2/(9^n) ?

The sum from n=0 to infinity of 2/(9^n) is 9/4

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sum from n=0 to infinity of 2/(9^n)

n = 0 \sum \infty \frac{2}{9 ^{n}}

\frac{9}{4}

Solution steps

n = 0 \sum \infty \frac{2}{9 ^{n}}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty \frac{1}{9 ^{n}}

Apply Series Geometric Test:

\frac{9}{8}

= 2 \cdot \frac{9}{8}

Simplify

2 \cdot \frac{9}{8} : \frac{9}{4}

= \frac{9}{4}

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

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

n = 1 \sum \infty 8 ln (\frac{n}{n + 1})

n = 0 \sum \infty 5 \cdot (\frac{1}{4})^{n}

n = 0 \sum \infty \frac{n}{n !}

What is the sum from n=0 to infinity of 2/(9^n) ?
The sum from n=0 to infinity of 2/(9^n) is 9/4