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

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

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sum from n=0 to infinity of-(-1)^{n+1}

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Divergence Test:

diverges

= - diverges

= diverges

\frac{d}{d x} (\frac{3}{4 x ^{2}})

f (x) = x^{4} - 13 x^{2} + 36

\int_{0}^{1} \frac{45}{x ^{5}} d x

(1 + \frac{4}{x})^{x}

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

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