Find the upper and lower limits, the supremum and infimum of {a_n} defined by

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

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 $\sqrt{2n+1}+\sqrt{2n}$. – André Nicolas May 26 '12 at 17:43

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, $\,\displaystyle{\overline{\lim_{n\to\infty}}a_n=\lim_{n\to\infty}\left(\sqrt{2n+1}-\sqrt{2n}\right)}=\lim_{n\to\infty}\frac{1}{\sqrt{2n+1}+\sqrt{2n}}=0$
@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
@Daniela Oh, I see...well, then yes: begin from $\,n=0\,$...:) – DonAntonio May 26 '12 at 19:09