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

The sum from n=0 to infinity of 3/(-3^n) is -9/2

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

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

- \frac{9}{2}

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

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

Apply Series Geometric Test:

\frac{3}{2}

= (- 3) \frac{3}{2}

Simplify

(- 3) \frac{3}{2} : - \frac{9}{2}

= - \frac{9}{2}

n = 1 \sum \infty \frac{5 ^{n}}{n + 1}

k = 1 \sum \infty \frac{6}{k}

n = 1 \sum \infty \frac{1}{3 n ^{2} + 9}

n = 1 \sum \infty \frac{1}{3 n ^{2} + 4}

n = 1 \sum \infty \frac{7}{n ^{9}}

What is the sum from n=0 to infinity of 3/(-3^n) ?
The sum from n=0 to infinity of 3/(-3^n) is -9/2