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

The sum from n=1 to infinity of (1)^{n-1} is diverges

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sum from n=1 to infinity of (1)^{n-1}

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

diverges

Solution steps

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

Simplify

= n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= diverges

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

k = 2 \sum \infty (\frac{1}{4})^{2 k}

n = 1 \sum \infty \frac{9 n}{6 n + 1}

0 = 1 \sum \infty \frac{ln ( 0 )}{0}

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

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