What is the sum from n=1 to infinity of 8/((-3)^n) ?

The sum from n=1 to infinity of 8/((-3)^n) is converges

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sum from n=1 to infinity of 8/((-3)^n)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 8 converges

= converges

n = 2 \sum \infty \frac{1}{n ln ( n ^{2} )}

n = 5 \sum \infty 2 (- \frac{1}{4})^{n}

n = 2 \sum \infty \frac{1}{n ^{2} - n}

n = 1 \sum \infty \frac{1}{8 n + 4}

n = 1 \sum \infty \frac{3 ^{n} + n}{n !}

What is the sum from n=1 to infinity of 8/((-3)^n) ?
The sum from n=1 to infinity of 8/((-3)^n) is converges