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}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (9 x - 5)

\int (x - 8)^{\frac{3}{2}} d x

\frac{d}{d x} (e^{2 x y})

f (x) = x^{2} (1 - 4 x)

f (x) = x^{3} - 6 x^{2} - 34 x + 40, at x = - 3

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