What is the sum from n=2 to infinity of (3/7)^{4n} ?

The sum from n=2 to infinity of (3/7)^{4n} is converges

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sum from n=2 to infinity of (3/7)^{4n}

n = 2 \sum \infty (\frac{3}{7})^{4 n}

converges

Solution steps

n = 2 \sum \infty (\frac{3}{7})^{4 n}

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty \frac{6 n}{4 n + 1}

n = 1 \sum \infty \frac{2}{5 n ^{2} + 5 n}

n = 0 \sum \infty (\frac{5}{12})^{n}

n = 0 \sum \infty \frac{3}{5 ^{n}} + \frac{2}{n}

What is the sum from n=2 to infinity of (3/7)^{4n} ?
The sum from n=2 to infinity of (3/7)^{4n} is converges