What is the sum from n=1 to infinity of 7(0.8)^{n-1} ?

The sum from n=1 to infinity of 7(0.8)^{n-1} is converges

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sum from n=1 to infinity of 7(0.8)^{n-1}

n = 1 \sum \infty 7 (0.8)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 7 \cdot 0. 8^{n - 1}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty 0. 8^{n - 1}

Apply Series Ratio Test:

converges

= 7 converges

= converges

n = 1 \sum \infty \frac{n}{13 n + 4}

n = 0 \sum \infty \frac{4}{5 - n}

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

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

n = 0 \sum \infty (1 + \frac{2}{5 n})^{2}

What is the sum from n=1 to infinity of 7(0.8)^{n-1} ?
The sum from n=1 to infinity of 7(0.8)^{n-1} is converges