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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 4 converges

= converges

k = 1 \sum \infty \frac{k - 5}{k + 2}

n = 0 \sum \infty \frac{1}{n \cdot ( n + 1 )}

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

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

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

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