Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

"Let $x \in (0,1)$. Compute with careful proof the lowest upper bound and greatest lower bound of $x^n$, where $n \in\mathbb{N} $."

From my understanding, I would say the lowest upper bound is $1$. Clearly, $1$ is an upper bound of $x^n$ since numbers that are less than $1$ raised to a power greater than $1$ decrease in size. However, how would go about proving that it is indeed $1$?

As for the greatest lower bound, I would say it would be $0$. Clearly it is a lower bound since $0$ to a power is always $0$, which is less than the possible outcomes for $x$. However, I also remember that as n approaches infinity, the limit of $0^0$ approaches $1$, so that has me a bit confused on whether $0$ is the greatest lower bound I'm looking for.

share|improve this question

1 Answer 1

It seems clear that $x$ is intended to be a fixed number in the interval $(0,1)$.

The answer to the least upper bound question depends on how one defines $\mathbb{N}$. If $0$ is included in $\mathbb{N}$, then the least upper bound is $1$. If $0$ is not included, then the least upper bound is $x$. That will be easy to prove, since $x^n\lt x^1$ for every $n\ge 2$. So the least upper bound is in fact a maximum.

The greatest lower bound is $0$, but not for the reason you described. Certainly $0\lt x^n$ for every $n\in \mathbb{N}$, which makes $0$ a lower bound. We want to show that there is no lower bound greater than $0$. So you need to show that for any $b\gt 0$, there is an $n$ such that $x^n \lt b$.

To do this, I suggest rewriting $x$ as $\frac{1}{1+y}$, where $y$ is positive. We want to show that for some $n$, we have $\frac{1}{(1+y)^n}\lt b$, or equivalently that there is an $n$ such that $(1+y)^n \gt 1/b$.

You can prove that result by observing that $(1 + y)^n \ge 1 + ny$ for $y \ge 0$ (It is true for $y \gt -1$.) This is the Bernoulli Inequality, which you may already have met. The Inequality can be proved by induction. Or else you can show using the Binomial Theorem that $(1 + y)^n \ge 1 + ny$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.