What is the sum from n=0 to infinity of (-2)^n ?

The sum from n=0 to infinity of (-2)^n is diverges

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sum from n=0 to infinity of (-2)^n

n = 0 \sum \infty (- 2)^{n}

diverges

Solution steps

n = 0 \sum \infty (- 2)^{n}

Apply Series Geometric Test:

diverges

= diverges

\frac{\partial}{\partial x} (cos (4 y) e^{x})

x \to 2 lim (\frac{5 x - 25}{x - 2})

x^{^{'}} = - 0.0002 x

x \to 0 lim (\frac{3 x}{2 x ^{2} + x})

\int (e^{4 x} - 4)^{\frac{1}{5}} \cdot 8 e^{4 x} d x

What is the sum from n=0 to infinity of (-2)^n ?
The sum from n=0 to infinity of (-2)^n is diverges