# Math Help - Sequence related proof

1. ## Sequence related proof

Suppose that $\{ X_{n}\}_{n=1}^{\infty}$ is a sequence of real numbers that converges to $x_{0}$ and that all $x_{n}$ and $x_{0}$ are non-zero.

(i) Prove that there is a positive number $B$ that $|x_{n}|\geq B$ for all n.

(ii)Using part (i), prove that $\{\frac{1}{x_{n}}\}$ converges to $\{\frac{1}{x_{0}}\}$

OK, so here is how I did part (i):

Let $|x_{s}|$ be the smallest absolute value in the given sequence, i.e. $|x_{s}|=min(|x_{n}|) \forall n$
For $\epsilon\geq 0$,
In $N_{\epsilon}(|x_{s}|)$ there exists a number y such that $y\leq x_{s}\to y\leq |x_{n}|$ for all n.

(ii) [I have no idea how to do this one -- or even that the first part is right. So any help will be greatly appreciated. Thanks in advance]

2. I'll give a couple of very nitpicking remarks.

Originally Posted by ashimb9
Let $|x_{s}|$ be the smallest absolute value in the given sequence
When one says "let x be y", usually x is a fresh variable, a new name for the described object y. If x is not just a name, but an expression, then it is not clear that the right-hand side y can have the form x. For example, consider "let (x+2)(x-1) be -2". Does such x exist, and if it does, is it unique? So, wrapping $x_s$ in the absolute value function, as in your case, is suspicious from the start.

One should say, "Let s be such that $|x_s|\le |x_n|$ for all n". Here, again, one has to describe why such s exists. And, in fact, it does not have to: consider $x_n=1/n$. In general, a set of real numbers bounded from below always has an infimum but does not always have a minimum (which by definition has to be an element of the set). This is an essential point of this problem.

For $\epsilon\geq 0$,
In $N_{\epsilon}(|x_{s}|)$ there exists a number y such that $y\leq x_{s}\to y\leq |x_{n}|$ for all n.
It's not clear to me what $N_{\epsilon}(|x_{s}|)$ is. Also, if such y exists, why is it positive, as required by the question?

A correct way is to find N such that, say, $|x_n|>1/2|x_0|$ for all n > N. Then the initial segment $\{x_n\}_{n=0}^N$ is finite, and finite sets have not only infimum but minimum as well. Then you can find the positive lower bound for all $|x_n|$.

For (ii), try to find an upper bound on $|1/x_n-1/x_0|$ in terms of $|x_n-x_0|$, B, and $|x_0|$.