What is the sum from n=1 to infinity of 4^{-3n} ?

The sum from n=1 to infinity of 4^{-3n} is converges

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sum from n=1 to infinity of 4^{-3n}

n = 1 \sum \infty 4^{- 3 n}

converges

Solution steps

n = 1 \sum \infty 4^{- 3 n}

Apply Series Ratio Test:

converges

= converges

k = 0 \sum \infty e^{- k}

n = 0 \sum \infty \frac{3}{5 - n}

n = 2 \sum \infty \frac{4}{n ln ( n )}

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

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

What is the sum from n=1 to infinity of 4^{-3n} ?
The sum from n=1 to infinity of 4^{-3n} is converges