What is the sum from n=0 to infinity of (-1)^n 1/2 ?

The sum from n=0 to infinity of (-1)^n 1/2 is diverges

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sum from n=0 to infinity of (-1)^n 1/2

n = 0 \sum \infty (- 1)^{n} \frac{1}{2}

diverges

Solution steps

n = 0 \sum \infty (- 1)^{n} \frac{1}{2}

Apply the constant multiplication rule:

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

= \frac{1}{2} \cdot n = 0 \sum \infty (- 1)^{n}

Apply Series Geometric Test:

diverges

= \frac{1}{2} diverges

= diverges

n = 1 \sum \infty \frac{7}{5 n - 1}

n = 2 \sum \infty (\frac{1}{8})^{n + 4}

n = 0 \sum \infty \frac{3 n}{1 + n ^{4}}

n = 1 \sum \infty 6 - 5 (0.7)^{n}

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

What is the sum from n=0 to infinity of (-1)^n 1/2 ?
The sum from n=0 to infinity of (-1)^n 1/2 is diverges