What is the sum from n=2 to infinity of (1/8)^{n+4} ?

The sum from n=2 to infinity of (1/8)^{n+4} is converges

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sum from n=2 to infinity of (1/8)^{n+4}

n = 2 \sum \infty (\frac{1}{8})^{n + 4}

converges

Solution steps

n = 2 \sum \infty (\frac{1}{8})^{n + 4}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{3 n}{1 + n ^{4}}

n = 1 \sum \infty 6 - 5 (0.7)^{n}

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

n = 0 \sum \infty (sin (3))^{n}

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

What is the sum from n=2 to infinity of (1/8)^{n+4} ?
The sum from n=2 to infinity of (1/8)^{n+4} is converges