# Sequence

#### godai

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

#### Sudharaka

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).

#### igopogo

...and a bounded above sequence could be a similar monotonic decreasing sequence.

#### matheagle

MHF Hall of Honor
-1,-2,-3,-4,... likewise goes to negative infinity and so does all it subsequences.

#### godai

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?

#### igopogo

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
(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 $$\displaystyle x_n = \left\{ {\begin{array}{rl} {n,} & {even} \\ { - n,} & {odd} \\ \end{array} } \right.$$

#### godai

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.
its limit is e and its not rational right?
but is it the sequence rational?

#### Drexel28

MHF Hall of Honor
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.
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
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.