What is the sum from n=1 to infinity of 4(0.9)^{n-1} ?

The sum from n=1 to infinity of 4(0.9)^{n-1} is converges

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sum from n=1 to infinity of 4(0.9)^{n-1}

n = 1 \sum \infty 4 (0.9)^{n - 1}

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 4 converges

= converges

n = 10 \sum \infty \frac{3}{n ^{4.6}}

n = 1 \sum \infty \frac{1}{2 n + 5}

n = 0 \sum \infty \frac{n}{n ^{n}}

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

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

What is the sum from n=1 to infinity of 4(0.9)^{n-1} ?
The sum from n=1 to infinity of 4(0.9)^{n-1} is converges