What is the sum from n=2 to infinity of (1/4)^{3n} ?

The sum from n=2 to infinity of (1/4)^{3n} is converges

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sum from n=2 to infinity of (1/4)^{3n}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty (n - 1) (\frac{1}{2})^{n}

n = 1 \sum \infty \frac{n + 1}{n + 3}

n = 0 \sum \infty \frac{1}{4 n ^{2}}

n = 0 \sum \infty 8 \cdot (\frac{2}{7})^{n}

k = 1 \sum \infty k e^{- 7 k^{2}}

What is the sum from n=2 to infinity of (1/4)^{3n} ?
The sum from n=2 to infinity of (1/4)^{3n} is converges