What is the sum from n=0 to infinity of 2((-2)/3)^n ?

The sum from n=0 to infinity of 2((-2)/3)^n is 6/5

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sum from n=0 to infinity of 2((-2)/3)^n

n = 0 \sum \infty 2 (\frac{- 2}{3})^{n}

\frac{6}{5}

Solution steps

n = 0 \sum \infty 2 (\frac{- 2}{3})^{n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty (\frac{- 2}{3})^{n}

Apply Series Geometric Test:

\frac{3}{5}

= 2 \cdot \frac{3}{5}

Simplify

2 \cdot \frac{3}{5} : \frac{6}{5}

= \frac{6}{5}

\frac{d y}{d x} = e^{- (x + y)}

\int e^{- x + 2} d x

y = 4 (x + 1), y = 5 (x + 1), x = 5

f (x) = 9 sin (x) cos (x)

\int sin (x) cos (2 x) d x

What is the sum from n=0 to infinity of 2((-2)/3)^n ?
The sum from n=0 to infinity of 2((-2)/3)^n is 6/5