What is the sum from n=0 to infinity of 3(-1)^n ?

The sum from n=0 to infinity of 3(-1)^n is diverges

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sum from n=0 to infinity of 3(-1)^n

n = 0 \sum \infty 3 (- 1)^{n}

diverges

Solution steps

n = 0 \sum \infty 3 (- 1)^{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty (- 1)^{n}

Apply Series Geometric Test:

diverges

= 3 diverges

= diverges

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

n = 3 \sum \infty \frac{4}{n ^{2} - 4}

k = 1 \sum \infty \frac{1}{k + 3}

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

n = 2 \sum \infty \frac{6}{3 ^{2 n}}

What is the sum from n=0 to infinity of 3(-1)^n ?
The sum from n=0 to infinity of 3(-1)^n is diverges