What is the sum from n=1 to infinity of 8(0.3)^{n-1} ?

The sum from n=1 to infinity of 8(0.3)^{n-1} is converges

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sum from n=1 to infinity of 8(0.3)^{n-1}

n = 1 \sum \infty 8 (0.3)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 8 \cdot 0. 3^{n - 1}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty 0. 3^{n - 1}

Apply Series Ratio Test:

converges

= 8 converges

= converges

n = 0 \sum \infty \frac{10}{n ^{2} + 9}

b = 0 \sum \infty 9 (\frac{2}{3})^{b}

n = 0 \sum \infty \frac{( - 2 ) ^{n}}{n}

n = 1 \sum \infty \frac{n}{5 n ^{3} + 3}

n = 1 \sum \infty n e^{- 15 n}

What is the sum from n=1 to infinity of 8(0.3)^{n-1} ?
The sum from n=1 to infinity of 8(0.3)^{n-1} is converges