What is the sum from n=1 to infinity of 3(0.4)^{n-1} ?

The sum from n=1 to infinity of 3(0.4)^{n-1} is converges

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sum from n=1 to infinity of 3(0.4)^{n-1}

n = 1 \sum \infty 3 (0.4)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 3 \cdot 0. 4^{n - 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty 0. 4^{n - 1}

Apply Series Ratio Test:

converges

= 3 converges

= converges

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

n = 0 \sum \infty - (\frac{7}{3 ^{n}})

n = 1 \sum \infty \frac{8}{- 3 ^{n}}

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

n = 0 \sum \infty \frac{3 ^{n + 1}}{π}

What is the sum from n=1 to infinity of 3(0.4)^{n-1} ?
The sum from n=1 to infinity of 3(0.4)^{n-1} is converges