What is the sum from n=0 to infinity of 1/5*(-2/3)^n ?

The sum from n=0 to infinity of 1/5*(-2/3)^n is 3/25

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sum from n=0 to infinity of 1/5*(-2/3)^n

n = 0 \sum \infty \frac{1}{5} \cdot (- \frac{2}{3})^{n}

\frac{3}{25}

Solution steps

n = 0 \sum \infty \frac{1}{5} (- \frac{2}{3})^{n}

Apply the constant multiplication rule:

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

= \frac{1}{5} \cdot n = 0 \sum \infty (- \frac{2}{3})^{n}

Apply Series Geometric Test:

\frac{3}{5}

= \frac{1}{5} \cdot \frac{3}{5}

Simplify

\frac{1}{5} \cdot \frac{3}{5} : \frac{3}{25}

= \frac{3}{25}

x^{2} + y^{2} = 25, (- 4, 3)

f (x) = \frac{4 x}{x ^{2} + 3}, at x = 1

x \to 0 + lim (cos (\frac{π}{x}))

\frac{d}{d x} ((1 + x)^{\frac{1}{3}})

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

What is the sum from n=0 to infinity of 1/5*(-2/3)^n ?
The sum from n=0 to infinity of 1/5*(-2/3)^n is 3/25