What is the sum from n=0 to infinity of 8(7/8)^n ?

The sum from n=0 to infinity of 8(7/8)^n is 64

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sum from n=0 to infinity of 8(7/8)^n

n = 0 \sum \infty 8 (\frac{7}{8})^{n}

64

Solution steps

n = 0 \sum \infty 8 (\frac{7}{8})^{n}

Apply the constant multiplication rule:

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

= 8 \cdot n = 0 \sum \infty (\frac{7}{8})^{n}

Apply Series Geometric Test:

8

= 8 \cdot 8

Simplify

= 64

n = 1 \sum \infty 5 n sin (\frac{4}{n})

k = 1 \sum \infty k e^{- 2 k^{2}}

k = 1 \sum \infty \frac{9 k ^{3}}{k !}

n = 0 \sum \infty \frac{1}{3 n ^{2} + 8}

n = 0 \sum \infty n \cdot 0. 5^{n}

What is the sum from n=0 to infinity of 8(7/8)^n ?
The sum from n=0 to infinity of 8(7/8)^n is 64