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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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    @Daniela, of course it can. Nothing wrong or cumbersome about that.2012-05-26