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.

I have read almost all the questions and answers regarding the theories, definitions, and properties of the $\operatorname{glb}$ and $\operatorname{lub}$ of a sequence, but I cannot find these values for the following sequence:

$\frac{n^2}{n + 1}$

I apologize for how basic this question is, but appreciate any help.

share|improve this question

3 Answers 3

up vote 3 down vote accepted

Notice that $\frac{n^2}{n+1}$ is approximately $n$, so you expect that the sequence has no finite least upper bound. More carefully,

$$\frac{n^2}{n+1}>\frac{n^2-1}{n+1}=\frac{(n+1)(n-1)}{n+1}=n-1\;,$$

which clearly diverges to $\infty$.

Now look at the first few terms, for $n=0,1,2$, and $3$: they are $0,\frac12,\frac43$, and $\frac94$. It appears that the sequence really is simply increasing; if so, its first term is its greatest lower bound. To check that it’s increasing, look at the difference between two consecutive terms:

$$\frac{(n+1)^2}{n+2}-\frac{n^2}{n+1}=\frac{(n+1)^3-n^2(n+2)}{(n+1)(n+2)}\;;$$

I’ll leave it to you to finish the algebra to verify that the last fraction is always positive.

share|improve this answer
    
Thank you very much for the thorough answer; it is appreciated! –  David Oct 1 '12 at 11:54
1  
@David: You’re very welcome. –  Brian M. Scott Oct 1 '12 at 11:56

Evaluate it for some $n$'s and guess ($n=0,1,2,3..$)

What if $n\to\infty$?

share|improve this answer

For large $n$ the fraction is large and for small $n$ it is small. The smallest $n$ in $\mathbb N$ is $0$ so the smallest value of the sequence is zero. This is the greatest lower bound since it is a lower bound and itself a value of the sequence.

Now you need similar thinking to find the least upper bound. Is the sequence bounded?

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.