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

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

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

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

converges

Solution steps

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

Simplify

(- \frac{1}{2})^{n + 3} : (- \frac{1}{2})^{n} (- \frac{1}{2})^{3}

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

Apply the constant multiplication rule:

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

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

(- \frac{1}{2})^{3} = - \frac{1}{8}

= (- \frac{1}{8}) \cdot n = 1 \sum \infty (- \frac{1}{2})^{n}

Apply Series Ratio Test:

converges

= (- \frac{1}{8}) converges

= converges

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

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

k = 4 \sum \infty \frac{6}{k} - \frac{6}{k + 1}

n = 0 \sum \infty (\frac{3}{n})^{\frac{3}{n}}

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

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