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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{5 n}{6 n + 1}

n = 1 \sum \infty \frac{1}{6}

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

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

n = 1 \sum \infty 4 (0.5)^{n - 1}

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