What is the sum from k=4 to infinity of 3(-1/2)^k ?

The sum from k=4 to infinity of 3(-1/2)^k is converges

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sum from k=4 to infinity of 3(-1/2)^k

k = 4 \sum \infty 3 (- \frac{1}{2})^{k}

converges

Solution steps

k = 4 \sum \infty 3 (- \frac{1}{2})^{k}

Apply the constant multiplication rule:

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

= 3 \cdot k = 4 \sum \infty (- \frac{1}{2})^{k}

Apply Series Ratio Test:

converges

= 3 converges

= converges

What is the sum from k=4 to infinity of 3(-1/2)^k ?
The sum from k=4 to infinity of 3(-1/2)^k is converges