What is the sum from n=1 to infinity of (5/6)^{4n-5} ?

The sum from n=1 to infinity of (5/6)^{4n-5} is converges

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sum from n=1 to infinity of (5/6)^{4n-5}

n = 1 \sum \infty (\frac{5}{6})^{4 n - 5}

converges

Solution steps

n = 1 \sum \infty (\frac{5}{6})^{4 n - 5}

Simplify

(\frac{5}{6})^{4 n - 5} : (\frac{5}{6})^{4 n} (\frac{5}{6})^{- 5}

= n = 1 \sum \infty (\frac{5}{6})^{4 n} (\frac{5}{6})^{- 5}

Apply the constant multiplication rule:

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

= (\frac{5}{6})^{- 5} \cdot n = 1 \sum \infty (\frac{5}{6})^{4 n}

(\frac{5}{6})^{- 5} = \frac{7776}{3125}

= \frac{7776}{3125} \cdot n = 1 \sum \infty (\frac{5}{6})^{4 n}

Apply Series Ratio Test:

converges

= \frac{7776}{3125} converges

= converges

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

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

n = 1 \sum \infty (cos (6))^{n}

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

n = 0 \sum \infty 6^{- n}

What is the sum from n=1 to infinity of (5/6)^{4n-5} ?
The sum from n=1 to infinity of (5/6)^{4n-5} is converges