What is the sum from n=4 to infinity of 6^{-n} ?

The sum from n=4 to infinity of 6^{-n} is converges

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sum from n=4 to infinity of 6^{-n}

n = 4 \sum \infty 6^{- n}

converges

Solution steps

n = 4 \sum \infty 6^{- n}

Simplify

6^{- n} : \frac{1}{6 ^{n}}

= n = 4 \sum \infty \frac{1}{6 ^{n}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty n \cdot 2

n = 1 \sum \infty \frac{1}{3 n ^{5} + 7}

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

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

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

What is the sum from n=4 to infinity of 6^{-n} ?
The sum from n=4 to infinity of 6^{-n} is converges