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

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?

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    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 $\sqrt{2n+1}+\sqrt{2n}$.2012-05-26

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

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    Why you can not take the succession from n = 0?2012-05-26
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    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...2012-05-26
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    @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.2012-05-26
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    Because the book says that the sup = 12012-05-26
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    @Daniela Oh, I see...well, then yes: begin from $\,n=0\,$...:)2012-05-26
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    @Brian , I know that. I remarked that *usually* real/complex sequence are indexed from $n=1$ along the natural numbers. Of course, this doesn't mean ***always***.2012-05-26
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    @DonAntonio: I understood your statement. As I said, I’m not convinced that it’s true. And as a separate point, for me and many others, *indexed by the natural numbers* means starting from $0$, not $1$.2012-05-26
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    I believe that the sequence can start from 0 as long as it defined in the sequence2012-05-26
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    @Brian, We'll have to agree to disagree, and I won't post anymore on this. I just will direct you to Rudin's "Principles of Mathematical Analysis", definition 2.7, or chapter 26 in Spivak's "Calculus" , or to 1.1 in Lang's "Real and Functional Analysis", or definition 2.12 in Apostol's "Mathematical Analysis", or chapter 10.1 of Thomas's "Calculus", or chapter 11 of Swokowski's "Calculs with Analytic Geometry", to mention a few of the best well known books on the subject. In fact, I can't remember right now one single instance where the def. begins at n = 0 but, of course, there must be some.2012-05-26
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    @Daniela, of course it can. Nothing wrong or cumbersome about that.2012-05-26