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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

diverges

= i diverges

= diverges

(x + x) \frac{d y}{d x} = y + y

\frac{d}{d x} (4 arccos (1 - x^{2}))

\frac{\partial}{\partial y} (- 2 y^{2})

\int_{- 2}^{3} \frac{39}{x ^{4}} d x

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