What is the sum from n=2 to infinity of (1/3)^{2n} ?

The sum from n=2 to infinity of (1/3)^{2n} is converges

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sum from n=2 to infinity of (1/3)^{2n}

n = 2 \sum \infty (\frac{1}{3})^{2 n}

converges

Solution steps

n = 2 \sum \infty (\frac{1}{3})^{2 n}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{1}{n ^{e}}

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n - 3}

n = 1 \sum \infty \frac{1}{n ^{0.6}}

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

n = 4 \sum \infty n e^{- n^{2}}

What is the sum from n=2 to infinity of (1/3)^{2n} ?
The sum from n=2 to infinity of (1/3)^{2n} is converges