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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{1}{7 n - 1}

k = 1 \sum \infty k^{3} e^{- 2 k}

n = 1 \sum \infty (- 1.99 + 1)^{n}

n = 1 \sum \infty \frac{2}{n ^{10}}

n = 0 \sum \infty \frac{n ^{6}}{n !}

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