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 (3 x) d x

\int_{0}^{1} arcsin (y) d y

f (x) = x^{4} - 25 x^{2} + 144, at x = - 1

e^{\frac{x}{5}}

14 e^{x^{2}} x

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