Proving bounded monotonic sequences must converge

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SUMMARY

The discussion focuses on proving that bounded monotonic sequences must converge, utilizing the contrapositive approach. The key argument presented is that if a sequence {s(n)} does not converge, it must either be non-monotonic or unbounded. The proof demonstrates that if {s(n)} approaches +infinity, it cannot be bounded, thereby confirming that all monotonic sequences must converge. The proof's validity hinges on the application of contrapositives and the consideration of divergent cases.

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Homework Statement


I'm approaching this problem from a different method than conventially shown.


Homework Equations


if lim=infinity for all M>0, there exists a N such that n>N => {s(n)}>=M


The Attempt at a Solution



this can be rewritten as:

{s(n)} is a sequence. If {s(n)} is bounded and monotonic , then {s(n)} converges.

the contrapositive is,

{s(n)} is a sequence. If {s(n)} doesn't converge then it is either not monotonic or not bounded.


Hence, if I just show that it is not monotonic, then the proof works.


PROOF:

{s(n)} doesn't converge. So if {s(n)}=+infinity, then that means that for all M>0, there exists an N such for all n>N, M>={s(n)}.

Then {s(n)} is not bounded.

Thus all monotonic sequences must converge. Q.E.D



Does this proof work?

1)I'm a bit worried because I don't even know if I can apply contrapostives this way.

2)I don't know if I have to show {s(n)} is not bounded for all cases of {s(n)} diverging(i.e +infinity, -infinity, DNE). It seems plausible that if I can show that {s(n)} is not bounded when {s(n)} diverges for even just one of the cases, then the theorem works.
 
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With regard to your monotonicity statement and divergent sequences, take the sequence s_{n}=(-1)^{n}, this is bounded but not convergent, neither is it monotonic.

Mat
 

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