Sequence

May 2010
11
0
give counterexamples for the claims
1) every monotonic sequence has a convergent subsequence
2) every bounded above sequence has a convergent subsequence
3) every increasing sequence has a convergent subsequence
 
Dec 2009
872
381
1111
give counterexamples for the claims
1) every monotonic sequence has a convergent subsequence
2) every bounded above sequence has a convergent subsequence
3) every increasing sequence has a convergent subsequence
Dear godai,

Consider the sequence, \(\displaystyle S_{n}=n=\{1,2,3,4..........\}\)

This sequence does not have a convergent subsequence. If you take any subsequence it tends to infinity. This is a strictly increasing sequence. Hence it is also a monotonic sequence. Therefore this is a counterexample for parts (1) and (3).

Hope this will help you.
 
May 2010
9
0
...and a bounded above sequence could be a similar monotonic decreasing sequence.
 

matheagle

MHF Hall of Honor
Feb 2009
2,763
1,146
-1,-2,-3,-4,... likewise goes to negative infinity and so does all it subsequences.
 
May 2010
11
0
thanks :)

(a sequence not bounded above and doesn't tend to infinity)
was thinking about 0 for all n
it's not bounded above or below is it?
 
May 2010
9
0
Yes, it is bounded above and below by, say, 1 and -1. Play around with things that might add up to zero, though.
 

Plato

MHF Helper
Aug 2006
22,455
8,631
(a sequence not bounded above and doesn't tend to infinity)
was thinking about 0 for all n it's not bounded above or below is it?
Try \(\displaystyle x_n = \left\{ {\begin{array}{rl} {n,} & {even} \\
{ - n,} & {odd} \\ \end{array} } \right.\)
 
May 2010
11
0
hmm okay :) thanks for the replies! :)

unfortunately I have more questions...

say if a sequence a_n converes to a, and a_n > 1 for all n >=1, show that a>=1
It's obvious right? if it's a sequence say (n+1)/n, it's always bigger than 1 and it's limit is 1 but how do i show it? :(

a sequence (cos n )/ (2 - sin n ) is bounded right?
it does not converge or is monotone right? :)

just want to clarify :
the question is find a Cauchy sequence such that it is rational for all n bigger or equals to 1 such that its limit is not rational.
was thinking about [(n+1)/n]^n
its limit is e and its not rational right?
but is it the sequence rational?
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
1,566
Berkeley, California
hmm okay :) thanks for the replies! :)

unfortunately I have more questions...

say if a sequence a_n converes to a, and a_n > 1 for all n >=1, show that a>=1
It's obvious right? if it's a sequence say (n+1)/n, it's always bigger than 1 and it's limit is 1 but how do i show it? :(
You need to give us a little more background than that. What are you learning? What are your thoughts? My personal preference (most likely not your own) would be to note that \(\displaystyle a\) is either a limit point of \(\displaystyle (0,\infty)\) or a point of it. Pertinently, \(\displaystyle a\in\overline{(0,\infty)}=[0,\infty)\). Does that make sense? Does that look like what you're doing?

a sequence (cos n )/ (2 - sin n ) is bounded right?
it does not converge or is monotone right? :)
You tell us. \(\displaystyle \left|\frac{\cos(n)}{2-\sin(n)}\right|=\frac{|cos(n)|}{|2-\sin(n)|}\). Now, you want to make the numerator as big as possible (what is cosine always less than in absolute value?) and the denominator as small as possible (hint: \(\displaystyle |x|-|y|\leqslant |x-y|\))

just want to clarify :
the question is find a Cauchy sequence such that it is rational for all n bigger or equals to 1 such that its limit is not rational.
was thinking about [(n+1)/n]^n
its limit is e and its not rational right?
Correct!
but is it the sequence rational?
Of course. Each \(\displaystyle \frac{n+1}{n}\) is clearly rational and the rationals are closed under finite multiplication.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
hmm okay :) thanks for the replies! :)

unfortunately I have more questions...

say if a sequence a_n converes to a, and a_n > 1 for all n >=1, show that a>=1
It's obvious right? if it's a sequence say (n+1)/n, it's always bigger than 1 and it's limit is 1 but how do i show it? :(
By contradiction. If a< 1, let \(\displaystyle \epsilon= (1- a)/2\). Then we must have \(\displaystyle |a_n- a|< \epsilon\) for sufficiently large n which is impossible if \(\displaystyle a_n> 1\).

a sequence (cos n )/ (2 - sin n ) is bounded right?
Yes. \(\displaystyle -1\le cos n\le 1\) while \(\displaystyle 1\le 2- sin n\le 3\).

it does not converge or is monotone right? :)

just want to clarify :
the question is find a Cauchy sequence such that it is rational for all n bigger or equals to 1 such that its limit is not rational.
was thinking about [(n+1)/n]^n
its limit is e and its not rational right?
but is it the sequence rational?