What is the sum from n=2 to infinity of (3/7)^{3n} ?

The sum from n=2 to infinity of (3/7)^{3n} is converges

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sum from n=2 to infinity of (3/7)^{3n}

n = 2 \sum \infty (\frac{3}{7})^{3 n}

converges

Solution steps

n = 2 \sum \infty (\frac{3}{7})^{3 n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{2}{n ^{7}}

n = 1 \sum \infty \frac{1}{n ^{25}}

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

n = 1 \sum \infty 4 (- \frac{10}{11})^{n}

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

What is the sum from n=2 to infinity of (3/7)^{3n} ?
The sum from n=2 to infinity of (3/7)^{3n} is converges