What is the sum from n=0 to infinity of 1/2*(-1)^n ?

The sum from n=0 to infinity of 1/2*(-1)^n is diverges

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sum from n=0 to infinity of 1/2*(-1)^n

n = 0 \sum \infty \frac{1}{2} \cdot (- 1)^{n}

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= \frac{1}{2} \cdot n = 0 \sum \infty (- 1)^{n}

Apply Series Geometric Test:

diverges

= \frac{1}{2} diverges

= diverges

\frac{\partial}{\partial x} (- 4 x y - x^{4} - y^{4})

f (x) = x + 4

\int \frac{x ^{2} - x + 14}{x ^{3} + 7 x} d x

\int \frac{x ^{2} + 2 x}{x ^{3} + 3 x ^{2} + 2} d x

What is the sum from n=0 to infinity of 1/2*(-1)^n ?
The sum from n=0 to infinity of 1/2*(-1)^n is diverges