What is the sum from j=1 to infinity of 2^{-4j} ?

The sum from j=1 to infinity of 2^{-4j} is converges

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sum from j=1 to infinity of 2^{-4j}

j = 1 \sum \infty 2^{- 4 j}

converges

Solution steps

j = 1 \sum \infty 2^{- 4 j}

Apply Series Ratio Test:

converges

= converges

n = 3 \sum \infty (\frac{1}{3})^{n}

n = 1 \sum \infty n! (- e)^{- 2 n}

n = 0 \sum \infty e^{\frac{- n ^{6}}{8}}

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

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

What is the sum from j=1 to infinity of 2^{-4j} ?
The sum from j=1 to infinity of 2^{-4j} is converges