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

\int e^{- a x} sin (x) d x

(2 sec^{2} (x) tan (x))^{^{'}}

\int \frac{1}{( x + 2 ) x ^{2} + 4 x - 12} d x

\frac{1}{1 + x ^{2}}

x \to 0 - lim (6 x - 1)

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