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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty \frac{n ^{4}}{9 ^{n}}

n = 1 \sum \infty \frac{2 + sin ( n )}{n}

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

n = 1 \sum \infty 2 (0.7)^{n - 1}

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