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

The sum from n=0 to infinity of 5(-1/3)^n is 15/4

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

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

\frac{15}{4}

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

\frac{3}{4}

= 5 \cdot \frac{3}{4}

Simplify

5 \cdot \frac{3}{4} : \frac{15}{4}

= \frac{15}{4}

x \to 4 lim (\frac{( x ^{2} - 16 )}{2 - x})

\frac{d}{d x} (ln (2 e^{- x} sin (x)))

\int 1 + (tan (x))^{2} d x

g (x) = 6 - x

\frac{1}{1 - x}, 0, 1

What is the sum from n=0 to infinity of 5(-1/3)^n ?
The sum from n=0 to infinity of 5(-1/3)^n is 15/4