What is the sum from n=9 to infinity of 4/(8^n) ?

The sum from n=9 to infinity of 4/(8^n) is converges

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sum from n=9 to infinity of 4/(8^n)

n = 9 \sum \infty \frac{4}{8 ^{n}}

converges

Solution steps

n = 9 \sum \infty \frac{4}{8 ^{n}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 9 \sum \infty \frac{1}{8 ^{n}}

Apply Series Ratio Test:

converges

= 4 converges

= converges

n = 0 \sum \infty \frac{n ^{2}}{1}

k = 1 \sum \infty 4^{- k}

k = 0 \sum \infty \frac{1}{900 + 3 k}

n = 1 \sum \infty 6^{- 2 n}

n = 1 \sum \infty 3 (- \frac{1}{7})^{3 n}

What is the sum from n=9 to infinity of 4/(8^n) ?
The sum from n=9 to infinity of 4/(8^n) is converges