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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty (\frac{- 3}{4})^{n}

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

k = 2 \sum \infty 9 e^{- 2 k}

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

n = 1 \sum \infty (0)^{n - 1}

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