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

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

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sum from n=1 to infinity of (-1)^{2n}

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

diverges

Solution steps

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

(- 1)^{2 n} = 1

= n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= diverges

n = 0 \sum \infty \frac{1}{1 - n}

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

n = 1 \sum \infty \frac{5}{4 ^{n} - 3 ^{n}}

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

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

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