What is the sum from k=0 to infinity of-1^k ?

The sum from k=0 to infinity of-1^k is diverges

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sum from k=0 to infinity of-1^k

k = 0 \sum \infty - 1^{k}

diverges

Solution steps

k = 0 \sum \infty - 1^{k}

Apply the constant multiplication rule:

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

= - k = 0 \sum \infty 1^{k}

Simplify

= - k = 0 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= - diverges

= diverges

n = 1 \sum \infty \frac{5}{n ^{2} + n ^{3}}

n = 0 \sum \infty e^{- 1 n}

n = 1 \sum \infty \frac{2 n}{7 n - 6}

n = 1 \sum \infty n + 3

n = 0 \sum \infty \frac{7}{5 n + 4}

What is the sum from k=0 to infinity of-1^k ?
The sum from k=0 to infinity of-1^k is diverges