The sequence r^n tends to infinity for r > 1

Example

Let $r$ be a real number with $r > 1$. Then $r^n \to +\infty$ as $n \to \infty$, in the sense of a sequence tending to infinity. Write $r = 1 + h$ with $h > 0$. By Bernoulli's inequality, $r^n \ge 1 + n h$ for each natural number $n$. Let $M$ be a real number, and choose a natural number $N$ with $N h > M$. Then $r^n > M$ for all $n \ge N$.

Remarks

For $r < -1$ the sequence $r^n$ does not tend to $+\infty$ or to $-\infty$. Its terms change sign, but $|r^n| \to +\infty$. Exponential growth is faster than polynomial growth. The node exponentials dominate powers in sequences makes this precise.

Explanations

L1 human

Doubling again and again gives $1, 2, 4, 8, 16, \dots$, and these numbers grow without bound. Each factor $r > 1$ does the same, but a factor close to $1$ needs more steps.