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 8

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

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

8

Solution steps

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

x \to 0 lim (\frac{7 x + x ^{2}}{x})

\int 30 (25 - x^{2}) d x

f (x) = 7 ln (x), at x = 3

\frac{d}{d x} (a e^{x})

f (x) = (2 x + 1) (3 x - 2)

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 8