What is the sum from n=0 to infinity of 4(2/5)^n ?

The sum from n=0 to infinity of 4(2/5)^n is 20/3

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sum from n=0 to infinity of 4(2/5)^n

n = 0 \sum \infty 4 (\frac{2}{5})^{n}

\frac{20}{3}

Solution steps

n = 0 \sum \infty 4 (\frac{2}{5})^{n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 0 \sum \infty (\frac{2}{5})^{n}

Apply Series Geometric Test:

\frac{5}{3}

= 4 \cdot \frac{5}{3}

Simplify

4 \cdot \frac{5}{3} : \frac{20}{3}

= \frac{20}{3}

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

\frac{\partial}{\partial x} (- \frac{1}{4} (y^{2} + x^{2}))

\frac{\partial}{\partial x} (e^{- 3 t} sin (x + 5 t))

\int \frac{x ^{2}}{x ^{2} + 7} d x

x \to 2 lim (\frac{x ^{3}}{2})

What is the sum from n=0 to infinity of 4(2/5)^n ?
The sum from n=0 to infinity of 4(2/5)^n is 20/3