1) Let be a sequence such that converges. Clearly must have an infinite number of positive (or negative) terms. Take such that for all and then is a subsequence of , but converges.
2)Use the fact that the odd and even numbers partition
1) Can you find a sequnce of real nos. {sn} which has no convergent subsequence and yet for which {|sn|} converges.
I am not able to think of such a sequence.
2) If a subsequence of {sn} of terms with even subscripts converges to L, as well as the subsequence with odd subscripts (converges to L i.e.) then, prove that the given sequence converges to L.
Here I intuitively know the theorem statement is sort of obvious, but how do I go about the proof?
Any help will be appreciated.
I think for 2) more explicitly it is best to say:
Given , since the even and odd sequences converge to L, there should be an and an such that for all odd n bigger than you have and similarly for all even n bigger than you have .
Then let and you have for all that so the sequence converges to L as desired.
For the first question I don't think it's possible. For if were bounded, it would have a convergent subsequence by Bolzano-Weierstrass. Thus the sequence must be unbounded. But this too isn't possible because if converges we have that by (nothing is lost by assuming all terms positive really) that would converge. But is unbounded and thus diverges, and so must .