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.

$a_{2n}=-\sqrt{2n}$; $a_{2n+1}$=$\sqrt{2n+1}-\sqrt{2n}$

Lower limit and infimum both negative infinity as $-\sqrt{2n}$ gets arbitrarily small.

Upper limit = zero ($a_{2n+1}$ goes to zero as $n-> \infty$), supremum = 1 (obtained for n = 0)

Is it right to my procedure?

Is there a more formal method?

share|improve this question
    
The supremum is not at $n=0$. The sup of the sequence $(a_m)$ is reached at $m=1$. (Just a technicality!) For showing the upper limit, it would be more formally clear if you multiplied top and (missing) bottom by $\sqrt{2n+1}+\sqrt{2n}$. –  André Nicolas May 26 '12 at 17:43

1 Answer 1

up vote 2 down vote accepted

The supremum is a maximum and, as André said, it is obtained at $\,m=1\,:\,\,a_1=\sqrt{3}-\sqrt{2}\,$ , since the function $\,f(x):=\sqrt{2x+1}-\sqrt{2x}\,$ is monotone descending.

The infimum certainly is $\,-\infty\,$, as $\,-\sqrt{2n}\to -\infty\,$ , and since this is also the limit of a subseq. of the seq. this is also the $\,\displaystyle{\underline{\lim}_{n\to\infty}a_n}\,$ .

Finally, $\,\displaystyle{\overline{\lim_{n\to\infty}}a_n=\lim_{n\to\infty}\left(\sqrt{2n+1}-\sqrt{2n}\right)}=\lim_{n\to\infty}\frac{1}{\sqrt{2n+1}+\sqrt{2n}}=0$

share|improve this answer
    
Why you can not take the succession from n = 0? –  Daniela del Carmen May 26 '12 at 18:52
    
Of course you can, but that won't change the basic facts. Besides, sequences are usually indexed by the naturals which usually are considered to begin (at least wrt sequences) at 1... –  DonAntonio May 26 '12 at 18:55
    
@DonAntonio: Sequences are indexed from whatever point is convenient, and I’m not at all sure that $1$ is a more common starting point that $0$. The term natural number is ambiguous: for many of us it means the set of non-negative integers, not the set of positive integers. –  Brian M. Scott May 26 '12 at 19:05
    
Because the book says that the sup = 1 –  Daniela del Carmen May 26 '12 at 19:07
    
@Daniela Oh, I see...well, then yes: begin from $\,n=0\,$...:) –  DonAntonio May 26 '12 at 19:09

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.