Sequence-Constant Inequalities

**miatchguy** · April 15th 2010, 01:03 AM

If you have a sequence $x_{n}$ and an inequality
$(|x_{n}|)^{\frac{1}{2}}<\beta$

Does that mean that the point to which the sequence converges (if it does) is less than $\beta$ or does that mean $(|x_{n}|)^{\frac{1}{2}}<\beta, \forall n$ ?

**tonio** · April 15th 2010, 03:08 AM

Originally Posted by miatchguy

If you have a sequence $x_{n}$ and an inequality
$(|x_{n}|)^{\frac{1}{2}}<\beta$

Does that mean that the point to which the sequence converges (if it does) is less than $\beta$ or does that mean $(|x_{n}|)^{\frac{1}{2}}<\beta, \forall n$ ?

In a rather unsurprising fashion, $(|x_{n}|)^{\frac{1}{2}}<\beta$ means just $(|x_{n}|)^{\frac{1}{2}}<\beta$ . As simply as that.

They could tell you this inequality is true for any n or only for some (one, a few, infinite, all...) n's, but that's something you can't ask here but rather ask the person who gave you the inequality.

Now, if $x_n\xrightarrow [n\to\infty]{}\alpha$ then $|x_n|\xrightarrow [n\to\infty]{}|\alpha|$ , but nothing else can be said about the relation between $\alpha,\,\beta$ until we know more about the given inequality.

Tonio

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