What is the sum from n=0 to infinity of (-1^n)/(3^n) ?

The sum from n=0 to infinity of (-1^n)/(3^n) is converges

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sum from n=0 to infinity of (-1^n)/(3^n)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= - converges

= converges

n = 0 \sum \infty n^{- 9}

n = 1 \sum \infty (\frac{4}{9})^{n}

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

n = 1 \sum \infty (\frac{π}{e})^{n}

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

What is the sum from n=0 to infinity of (-1^n)/(3^n) ?
The sum from n=0 to infinity of (-1^n)/(3^n) is converges