What is the sum from k=1 to infinity of (1)^k ?

The sum from k=1 to infinity of (1)^k is diverges

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sum from k=1 to infinity of (1)^k

k = 1 \sum \infty (1)^{k}

diverges

Solution steps

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

Simplify

= k = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= diverges

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

n = 1 \sum \infty \frac{4}{7 n - 1}

n = 9 \sum \infty \frac{4}{8 ^{n}}

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

k = 1 \sum \infty 4^{- k}

What is the sum from k=1 to infinity of (1)^k ?
The sum from k=1 to infinity of (1)^k is diverges