What is the sum from n=2 to infinity of 2^{-4n} ?

The sum from n=2 to infinity of 2^{-4n} is converges

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sum from n=2 to infinity of 2^{-4n}

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

converges

Solution steps

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

Simplify

2^{- 4 n} : 1 6^{- n}

= n = 2 \sum \infty 1 6^{- n}

Apply Series Root Test:

converges

= converges

n = 0 \sum \infty \frac{n}{6 + n ^{4}}

n = 1 \sum \infty n^{5} e^{n^{6}}

n = 5 \sum \infty n^{- s i n (3)}

n = 1 \sum \infty \frac{5}{n ^{10}}

n = 0 \sum \infty \frac{7}{2} (- \frac{3}{4})^{n}

What is the sum from n=2 to infinity of 2^{-4n} ?
The sum from n=2 to infinity of 2^{-4n} is converges