What is the sum from n=7 to infinity of 2^{-n} ?

The sum from n=7 to infinity of 2^{-n} is converges

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sum from n=7 to infinity of 2^{-n}

n = 7 \sum \infty 2^{- n}

converges

Solution steps

n = 7 \sum \infty 2^{- n}

Simplify

2^{- n} : \frac{1}{2 ^{n}}

= n = 7 \sum \infty \frac{1}{2 ^{n}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty (\frac{1}{2})^{n - 1}

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

n = 0 \sum \infty (\frac{11}{12})^{n}

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

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

What is the sum from n=7 to infinity of 2^{-n} ?
The sum from n=7 to infinity of 2^{-n} is converges