What is the sum from n=1 to infinity of 9(0.4)^{n-1} ?

The sum from n=1 to infinity of 9(0.4)^{n-1} is converges

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sum from n=1 to infinity of 9(0.4)^{n-1}

n = 1 \sum \infty 9 (0.4)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 9 \cdot 0. 4^{n - 1}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty 0. 4^{n - 1}

Apply Series Ratio Test:

converges

= 9 converges

= converges

n = 1 \sum \infty \frac{2}{n ^{2} + 9}

k = 1 \sum \infty \frac{k ^{2}}{3 ^{k}}

i = 1 \sum \infty \frac{1}{4 i ^{2} - 1}

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

n = 1 \sum \infty cos (6)

What is the sum from n=1 to infinity of 9(0.4)^{n-1} ?
The sum from n=1 to infinity of 9(0.4)^{n-1} is converges