What is the sum from n=0 to infinity of (4/9)^n ?

The sum from n=0 to infinity of (4/9)^n is 9/5

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sum from n=0 to infinity of (4/9)^n

n = 0 \sum \infty (\frac{4}{9})^{n}

\frac{9}{5}

Solution steps

n = 0 \sum \infty (\frac{4}{9})^{n}

Apply Series Geometric Test:

\frac{9}{5}

= \frac{9}{5}

x^{2} - 8, 1

\frac{d y}{d x} = y

y^{^{''}} + y + x = 0, y (0) = 0, y^{^{'}} (1) = 0

\frac{s + 2}{s ^{2} + 4 s + 6}

\frac{d}{d x} ((\frac{2}{5})^{x})

What is the sum from n=0 to infinity of (4/9)^n ?
The sum from n=0 to infinity of (4/9)^n is 9/5