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

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

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

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

\frac{3}{4}

Solution steps

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

Simplify

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

Apply Series Geometric Test:

\frac{3}{4}

= \frac{3}{4}

\int 56 x d x

x \to 0 lim (\frac{1}{x + 2} - \frac{1}{2})

y = 4 x^{- 3}

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

(x, y) \to (- 1, 1) lim (x + y)

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