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

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

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sum from n=0 to infinity of 3(4/5)^n

n = 0 \sum \infty 3 (\frac{4}{5})^{n}

15

Solution steps

n = 0 \sum \infty 3 (\frac{4}{5})^{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty (\frac{4}{5})^{n}

Apply Series Geometric Test:

5

= 3 \cdot 5

Simplify

= 15

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

k = 1 \sum \infty \frac{k !}{( 2 k ) !}

n = 1 \sum \infty (\frac{- e}{π})^{n}

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

n = 3 \sum \infty \frac{n ^{n}}{n !}

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