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

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

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

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

diverges

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{( - 1 ) ^{n}} = (- 1)^{- n}

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

Apply Series Divergence Test:

diverges

= diverges

\frac{\partial}{\partial y} (3 x^{2} y + 4 x y^{2})

\int e^{x} 8 + e^{x} d x

y^{^{'}} = \frac{( e ^{(x - y)} )}{1 + e ^{x}}

\int e^{- \frac{x}{10}} d x

\frac{d}{d x} (20 - 2 x)

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