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

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

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

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

diverges

Solution steps

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

Simplify

(- 1)^{2 n - 1} : (- 1)^{2 n} (- 1)^{- 1}

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

Apply the constant multiplication rule:

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

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

(- 1)^{- 1} = - 1

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

(- 1)^{2 n} = 1

= (- 1) \cdot n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= (- 1) diverges

= diverges

n = 0 \sum \infty \frac{1}{n ^{2} - 4}

k = 1 \sum \infty \frac{1}{7 ^{k} + 7}

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

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

n = 0 \sum \infty \frac{1}{3 e ^{n}}

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