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

The sum from n=0 to infinity of (1)^n is diverges

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sum from n=0 to infinity of (1)^n

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

diverges

Solution steps

n = 0 \sum \infty 1^{n}

Simplify

= n = 0 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= diverges

f (x) = 3 x^{2} - 4

\frac{\partial}{\partial y} (yz ln (x y))

\frac{\partial}{\partial x} (sin (π (5 x - 6 y)))

x, \frac{1}{x + y}

(2 + x) y^{^{'}} = 3 y

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