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

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

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

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

converges

Solution steps

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

Simplify

4^{3 - 3 n} : 4^{3} \cdot 4^{- 3 n}

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Simplify

= 64 \cdot n = 1 \sum \infty 4^{- 3 n}

Apply Series Ratio Test:

converges

= 64 converges

= converges

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

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

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

n = 0 \sum \infty 1 - \frac{1}{n}

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

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