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

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

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

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

8

Solution steps

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

Apply exponent rule:

\frac{a ^{c}}{b ^{c}} = (\frac{a}{b})^{c}

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

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

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

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

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

Simplify

= 8

\int_{- \infty}^{- 1} \frac{1}{x ^{3}} d x

\frac{d}{d x} (1 - 2 sin (x))

f (x) = 6 e^{x}

h \to + 6 lim (\frac{4 h}{h - 6})

f (x) = 9 - x

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