What is the sum from n=1 to infinity of (0.9)^n ?

The sum from n=1 to infinity of (0.9)^n is 9

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sum from n=1 to infinity of (0.9)^n

n = 1 \sum \infty (0.9)^{n}

9

Solution steps

n = 1 \sum \infty 0. 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 0. 9^{n} - 0. 9^{0}

n = 0 \sum \infty 0. 9^{n} = 10

= 10 - 0. 9^{0}

Simplify

= 9

25 y^{^{''}} - 15 y^{^{'}} - 4 y = 0

f (x) = x^{3} (x^{4} - 1)

\frac{d}{d x} (\frac{- x ^{2} - 4}{( x ^{2} - 4 ) ^{2}})

(x^{2} + 4) y = 8

\int_{0.5}^{1.5} (\frac{1}{8} + \frac{3}{8} x) d x

What is the sum from n=1 to infinity of (0.9)^n ?
The sum from n=1 to infinity of (0.9)^n is 9