What is the sum from n=2 to infinity of (3/8)^{2n} ?

The sum from n=2 to infinity of (3/8)^{2n} is converges

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sum from n=2 to infinity of (3/8)^{2n}

n = 2 \sum \infty (\frac{3}{8})^{2 n}

converges

Solution steps

n = 2 \sum \infty (\frac{3}{8})^{2 n}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{7}{π ^{n}}

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

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

n = 1 \sum \infty arccos (\frac{1}{n})

n = 2 \sum \infty \frac{n}{3 ^{n}}

What is the sum from n=2 to infinity of (3/8)^{2n} ?
The sum from n=2 to infinity of (3/8)^{2n} is converges