What is the sum from n=0 to infinity of-(7/(3^n)) ?

The sum from n=0 to infinity of-(7/(3^n)) is -21/2

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sum from n=0 to infinity of-(7/(3^n))

n = 0 \sum \infty - (\frac{7}{3 ^{n}})

- \frac{21}{2}

Solution steps

n = 0 \sum \infty - \frac{7}{3 ^{n}}

Apply the constant multiplication rule:

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

= - 7 \cdot n = 0 \sum \infty \frac{1}{3 ^{n}}

Apply Series Geometric Test:

\frac{3}{2}

= - 7 \cdot \frac{3}{2}

Simplify

- 7 \cdot \frac{3}{2} : - \frac{21}{2}

= - \frac{21}{2}

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

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

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

n = 0 \sum \infty 8 ln (4 - n)

n = 1 \sum \infty (- \frac{5}{6})^{n - 1}

What is the sum from n=0 to infinity of-(7/(3^n)) ?
The sum from n=0 to infinity of-(7/(3^n)) is -21/2