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

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

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

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

1

Solution steps

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

Apply the constant multiplication rule:

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

= 8 \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}

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

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

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

Simplify

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

= 1

n = 1 \sum \infty \frac{1}{n + 7}

n = 1 \sum \infty \frac{5}{n ^{2} + n}

n = 0 \sum \infty \frac{2}{n} - \frac{2}{n + 1}

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

k = 0 \sum \infty (\frac{1}{5})^{k}

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