What is the sum from n=0 to infinity of 9((1-3)/4)^n ?

The sum from n=0 to infinity of 9((1-3)/4)^n is converges

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sum from n=0 to infinity of 9((1-3)/4)^n

n = 0 \sum \infty 9 (\frac{1 - 3}{4})^{n}

converges

Solution steps

n = 0 \sum \infty 9 (\frac{1 - 3}{4})^{n}

Apply the constant multiplication rule:

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

= 9 \cdot n = 0 \sum \infty (\frac{1 - 3}{4})^{n}

\frac{1 - 3}{4} = - \frac{1}{2}

= 9 \cdot n = 0 \sum \infty (- \frac{1}{2})^{n}

Apply Series Ratio Test:

converges

= 9 converges

= converges

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

\int_{0}^{1} (- x + 2) - (x^{2} - 1) d x

\int x^{3} e^{x} d x

(x^{3} + y^{3}) d x + 3 x y^{2} d y = 0

\frac{d}{d x} ((8 x + 7)^{2})

What is the sum from n=0 to infinity of 9((1-3)/4)^n ?
The sum from n=0 to infinity of 9((1-3)/4)^n is converges