What is the sum from n=1 to infinity of 3(-1/2)^n ?

The sum from n=1 to infinity of 3(-1/2)^n is converges

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sum from n=1 to infinity of 3(-1/2)^n

n = 1 \sum \infty 3 (- \frac{1}{2})^{n}

converges

Solution steps

n = 1 \sum \infty 3 (- \frac{1}{2})^{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty (- \frac{1}{2})^{n}

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 0 \sum \infty 5 n + 5

n = 0 \sum \infty \frac{2 n !}{n ^{n}}

j = 2 \sum \infty j sin (\frac{10}{j})

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

n = 0 \sum \infty \frac{4}{8 ^{n}}

What is the sum from n=1 to infinity of 3(-1/2)^n ?
The sum from n=1 to infinity of 3(-1/2)^n is converges