What is the sum from n=1 to infinity of 2/(9^n) ?

The sum from n=1 to infinity of 2/(9^n) is 1/4

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sum from n=1 to infinity of 2/(9^n)

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

\frac{1}{4}

Solution steps

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

Apply the constant multiplication rule:

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

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

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 2 (n = 0 \sum \infty \frac{1}{9 ^{n}} - \frac{1}{9 ^{0}})

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

= 2 (\frac{9}{8} - \frac{1}{9 ^{0}})

Simplify

2 (\frac{9}{8} - \frac{1}{9 ^{0}}) : \frac{1}{4}

= \frac{1}{4}

3 x - 2 x^{2}

x \to 0 lim ((e^{- \frac{1}{x}})^{x})

\frac{d y}{d t} + t y = t y^{2}

\int x e^{2 x y} d y

\int x \cdot \frac{1}{x} d x

What is the sum from n=1 to infinity of 2/(9^n) ?
The sum from n=1 to infinity of 2/(9^n) is 1/4