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

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

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

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

diverges

Solution steps

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

Simplify

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

Apply Series Geometric Test:

diverges

= diverges

n = 1 \sum \infty \frac{e ^{4 n}}{n !}

x = 1 \sum \infty x

k = 2 \sum \infty \frac{1}{k + 1} - \frac{1}{k}

n = 0 \sum \infty \frac{5}{2 ^{n - 1}}

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

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