how to show xn>0 need not imply lin xn>a

**alice8675309** · February 21st 2011, 07:47 AM

Let (x_n) be a convergent sequence, and let a,b $\in$ R.

Show that for all n, x_n>a ned not imply limx_n>a.

How do I show this when I have previously proven that If x_n>= a for all n, then lim x_n>=a. (this was another question that i had proven before hand) But how would I show it, when it contradicts the previously stated question.

**Prove It** · February 21st 2011, 07:53 AM

What about the sequence $\displaystyle \left\{\frac{1}{n}\right\}$ . This is $\displaystyle > -1 \forall n \in \mathbf{Z}^+$ but $\displaystyle \to 0$ as $\displaystyle n \to \infty$

**chisigma** · February 21st 2011, 07:59 AM

The demonstration that $\displaystyle \lim_{n \rightarrow \infty} x_{n} = a$ doesn't imply in any case that $\forall n$ is $x_{n}>a$ or $x_{n}<a$ [the convergence may be 'oscillating'...] but that is $\lim_{n \rightarrow \infty} |x_{n}-a|=0$ ...

Kind regards

$\chi$ $\sigma$

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