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

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

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sum from n=1 to infinity of 4/((-3)^n)

n = 1 \sum \infty \frac{4}{( - 3 ) ^{n}}

converges

Solution steps

n = 1 \sum \infty \frac{4}{( - 3 ) ^{n}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{( - 3 ) ^{n}}

Apply Series Ratio Test:

converges

= 4 converges

= converges

n = 0 \sum \infty \frac{2}{2 ^{n}}

n = 0 \sum \infty ln (1 - n)

k = 0 \sum \infty \frac{k}{k + 1}

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

n = 0 \sum \infty \frac{n}{n + 5}

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