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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Simplify

= - n = 1 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= - diverges

= diverges

\frac{d}{d x} (\frac{2 x ^{3} - 7}{x ^{2}})

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

\int (8 cos (x) + 9 sin (x)) d x

\int - \frac{1}{4} csc (8 t) d t

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

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