What is the sum from n=0 to infinity of 8/((-3)^n) ?

The sum from n=0 to infinity of 8/((-3)^n) is 6

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sum from n=0 to infinity of 8/((-3)^n)

n = 0 \sum \infty \frac{8}{( - 3 ) ^{n}}

6

Solution steps

n = 0 \sum \infty \frac{8}{( - 3 ) ^{n}}

Apply the constant multiplication rule:

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

= 8 \cdot n = 0 \sum \infty \frac{1}{( - 3 ) ^{n}}

Apply Series Geometric Test:

\frac{3}{4}

= 8 \cdot \frac{3}{4}

Simplify

8 \cdot \frac{3}{4} : 6

= 6

n = 1 \sum \infty n \cdot 0. 5^{n}

n = 1.3 \sum \infty e^{- 1.8 n}

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

n = 1 \sum \infty \frac{n}{1 ^{n}}

n = 1 \sum \infty \frac{ln ( 8 n )}{n}

What is the sum from n=0 to infinity of 8/((-3)^n) ?
The sum from n=0 to infinity of 8/((-3)^n) is 6