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

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

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

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

\frac{9}{4}

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{3}{4}

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

Simplify

3 \cdot \frac{3}{4} : \frac{9}{4}

= \frac{9}{4}

y = x^{3} - 2 x + 1, (4, 57)

f (x) = \frac{sin ( x )}{1 + cos ( x )}

\frac{d}{d x} (\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} - 1)

y^{^{'}} = \frac{x}{6 y}

\frac{d y}{d t} = 0.12 - \frac{6 y}{50 - 2 t}

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