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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 3 converges

= converges

\int 20 e^{x} d x

\frac{d}{d x} (\frac{12}{x ^{4}})

(x, y) \to (0, 0) lim (x y^{8})

\frac{d}{d x} (\frac{1}{x ^{2} + c})

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

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