What is the sum from n=1 to infinity of 3(0.5)^{n-1} ?

The sum from n=1 to infinity of 3(0.5)^{n-1} is converges

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sum from n=1 to infinity of 3(0.5)^{n-1}

n = 1 \sum \infty 3 (0.5)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 3 \cdot 0. 5^{n - 1}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty 0. 5^{n - 1}

Apply Series Ratio Test:

converges

= 3 converges

= converges

x = 1 \sum \infty \frac{1}{x}

n = 1 \sum \infty \frac{9}{n}

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

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

n = 1 \sum \infty \frac{e ^{n}}{n ^{6}}

What is the sum from n=1 to infinity of 3(0.5)^{n-1} ?
The sum from n=1 to infinity of 3(0.5)^{n-1} is converges