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

The sum from n=1 to infinity of 1-(-1)^n is 0

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

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

0

Solution steps

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

Multiply by the conjugate of

1 - (- 1)^{n} : \frac{0}{1 + ( - 1 ) ^{n}}

= n = 1 \sum \infty \frac{0}{1 + ( - 1 ) ^{n}}

Simplify

= n = 1 \sum \infty 0

\sum 0 = 0

= 0

n = 0 \sum \infty cos (\frac{18}{n})

n = 3 \sum \infty \frac{5}{n ^{3}}

n = 1 \sum \infty (- \frac{1}{5})^{n - 1}

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

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

What is the sum from n=1 to infinity of 1-(-1)^n ?
The sum from n=1 to infinity of 1-(-1)^n is 0