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

The sum from n=0 to infinity of 1/3*(1/3)^n is 1/2

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

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

\frac{1}{2}

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{3}{2}

= \frac{1}{3} \cdot \frac{3}{2}

Cross-cancel:

3

= \frac{1}{2}

\int x^{2} sin (z x) d x

y^{^{'}} + \frac{4}{x} y = x^{9}

x \to \infty lim (arctan (\frac{( 1 - x ^{2} )}{( 1 + x )}))

\frac{d y}{d x} = 0.067 y (1 - \frac{y}{2800})

f (x) = x^{3}, (1, 1)

What is the sum from n=0 to infinity of 1/3*(1/3)^n ?
The sum from n=0 to infinity of 1/3*(1/3)^n is 1/2