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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\frac{t ^{4} + t ^{5} - 1}{t ^{6}}

\int (4 x + 1)^{6} d x

y^{3} - (18 x + 8) + 3 x y^{2} y^{^{'}} = 0

\int \frac{1}{y ^{3} y ^{2} - 9} d y

\frac{d}{d θ} (sin (4 θ))

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