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{1}{3} x^{3} - 3 x^{2} - 16 x + 8

y^{^{'}} cot^{2} (x) + tan (y) = 0

\int_{- 3}^{3} x^{2} d x

y^{^{'}} = - 1 - 4 x + 2 y, y (0) = 5

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

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