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

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

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

n = 0 \sum \infty (\frac{4}{5})^{n}

5

Solution steps

n = 0 \sum \infty (\frac{4}{5})^{n}

Apply Series Geometric Test:

5

= 5

\int 18 t e^{t} d t

5 x^{2} y^{^{'}} = y^{^{'}} + 7 x e - y

\int x^{2} (1 - x)^{\frac{3}{2}} d x

t y^{^{'}} = 2 y - 3 t^{2} cos (4 t)

\frac{\partial}{\partial x} (\frac{1}{2} e^{2 x})

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