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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

= - n = 0 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= - diverges

= diverges

\int \frac{2 x + 1}{x ^{2} + 2 x + 122} d x

\frac{\partial}{\partial y} (x^{2} y + y)

\int_{0}^{u} v + x d v

(1 - t^{2}) y^{^{'}} - t^{2} y = (1 + t) \cdot (1 - t^{2})

3 x^{2} + y^{2} = 1, (\frac{1}{2}, \frac{- 1}{2})

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