What is the sum from n=1 to infinity of 6(0.8)^{n-1} ?

The sum from n=1 to infinity of 6(0.8)^{n-1} is converges

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sum from n=1 to infinity of 6(0.8)^{n-1}

n = 1 \sum \infty 6 (0.8)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 6 \cdot 0. 8^{n - 1}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty 0. 8^{n - 1}

Apply Series Ratio Test:

converges

= 6 converges

= converges

n = 1 \sum \infty (cos (21))^{n}

n = 1 \sum \infty e^{- \frac{n}{2}}

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

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

n = 1 \sum \infty \frac{2 n + 1}{3 n}

What is the sum from n=1 to infinity of 6(0.8)^{n-1} ?
The sum from n=1 to infinity of 6(0.8)^{n-1} is converges