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[SOLVED] Convergence of Sequence

OhMyMarkov

Member
Mar 5, 2012
83
Hello everyone!

I am told that the limit of $\frac{x_{n+1}}{x_n}$ is $L>1$. I am asked to show that $\{x_n\}$ is not bounded and hence not convergent.

This is what I got so far:
Fix $\epsilon > 0$, $\exists n_0 \in N$ s.t. $\forall n > n_0$, we have
$|\frac{x_{n+1}}{x_n}-L|<\epsilon$.

Rearranging terms, we have:
$(L-\epsilon)\cdot x_n<x_{n+1}$

I'm stuck here, I want to show that there is an $M$ s.t. $x_n>M \; \forall n>n_1$ for some $n_1$. Thus the sequence is unbounded and converges to infinity. What I am thinking about is the following:
(1) Show that the sequence is increasing, then I get $x_n$ > $x_{n_1}$ which is a constant, but I don't know how.
(2) ... but then I don't know what to do with $\epsilon$ because it could be potentially bigger than $L$.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hello everyone!

I am told that the limit of $\frac{x_{n+1}}{x_n}$ is $L>1$. I am asked to show that $\{x_n\}$ is not bounded and hence not convergent.

This is what I got so far:
Fix $\epsilon > 0$, $\exists n_0 \in N$ s.t. $\forall n > n_0$, we have
$|\frac{x_{n+1}}{x_n}-L|<\epsilon$.

Rearranging terms, we have:
$(L-\epsilon)\cdot x_n<x_{n+1}$

I'm stuck here, I want to show that there is an $M$ s.t. $x_n>M \; \forall n>n_1$ for some $n_1$. Thus the sequence is unbounded and converges to infinity. What I am thinking about is the following:
(1) Show that the sequence is increasing, then I get $x_n$ > $x_{n_1}$ which is a constant, but I don't know how.
(2) ... but then I don't know what to do with $\epsilon$ because it could be potentially bigger than $L$.
Choose \(\varepsilon = (L+1)/2\) then there exists a \(n_0\) such that for all \(n>n_0\) we have:
\[\frac{x_{n+1}}{x_{n}} > L-\varepsilon>1\]

CB
 
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OhMyMarkov

Member
Mar 5, 2012
83
Hello CaptainBlack, thank you for your reply!

I tried choosing an $\epsilon = L-1$ so that there is an $n_0$ for which all $n>n_0$, we have
$|\frac{x_{n+1}}{x_n}-L|<L-1$
or
$x_n<x_{n+1}<(2L-1)\cdot x_n$

I get $\forall n > n_0$, $x_n$ is increasing.

And I just remembered that I should show that $x_n>M$ for all possible $M$, not for some $M$.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello everyone!

I am told that the limit of $\frac{x_{n+1}}{x_n}$ is $L>1$. I am asked to show that $\{x_n\}$ is not bounded and hence not convergent.

This is what I got so far:
Fix $\epsilon > 0$, $\exists n_0 \in N$ s.t. $\forall n > n_0$, we have
$|\frac{x_{n+1}}{x_n}-L|<\epsilon$.

Rearranging terms, we have:
$(L-\epsilon)\cdot x_n<x_{n+1}$

I'm stuck here, I want to show that there is an $M$ s.t. $x_n>M \; \forall n>n_1$ for some $n_1$. Thus the sequence is unbounded and converges to infinity. What I am thinking about is the following:
(1) Show that the sequence is increasing, then I get $x_n$ > $x_{n_1}$ which is a constant, but I don't know how.
(2) ... but then I don't know what to do with $\epsilon$ because it could be potentially bigger than $L$.
Let's consider the sequence $\lambda_{n}= \ln x_{n}$ so that is...

$\displaystyle \Delta_{n}= \lambda_{n+1}-\lambda_{n}= f(\lambda_{n})$ (1)

The sequence $\lambda_{n}$ converges to some finite limit $\Lambda$ only if $\lim_{n \rightarrow \infty} \Delta_{n}= 0$ but Your hypothesis says that that limit is >0, so that the sequence $\lambda_{n}$ diverges...

Kind regards

$\chi$ $\sigma$
 

OhMyMarkov

Member
Mar 5, 2012
83
Thanks for the reply.

I can see that the sequence converges, I have no problem with that. Moreover, I cannot use $\ln n$, we haven't covered logarithms yet, and won't be covering them probably.

Now I'm pretty sure there is a simple proof for this (perhaps a proof using Cauchy sequences).
 

chisigma

Well-known member
Feb 13, 2012
1,704
... I cannot use $\ln x_{n}$, we haven't covered logarithms yet, and won't be covering them probably...
I wonder how one can threat problems of Analysis with no knowledge for the present and also for the future of the concept of logarithm... a great mistery!...


Kind regards


$\chi$ $\sigma$
 

OhMyMarkov

Member
Mar 5, 2012
83
I wonder how one can threat problems of Analysis with no knowledge for the present and also for the future of the concept of logarithm... a great mistery!...


Kind regards


$\chi$ $\sigma$
I really appreciate your help by all means. It's not that I don't know about logarithm (but I very much do), it's that I was looking for something besides it in my proof.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hello CaptainBlack, thank you for your reply!

I tried choosing an $\epsilon = L-1$ so that there is an $n_0$ for which all $n>n_0$, we have
$|\frac{x_{n+1}}{x_n}-L|<L-1$
or
$x_n<x_{n+1}<(2L-1)\cdot x_n$

I get $\forall n > n_0$, $x_n$ is increasing.

And I just remembered that I should show that $x_n>M$ for all possible $M$, not for some $M$.
That the sequence is increasing is not sufficient to prove it unbounded.

CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Let's consider the sequence $\lambda_{n}= \ln x_{n}$ so that is...

$\displaystyle \Delta_{n}= \lambda_{n+1}-\lambda_{n}= f(\lambda_{n})$ (1)

The sequence $\lambda_{n}$ converges to some finite limit $\Lambda$ only if $\lim_{n \rightarrow \infty} \Delta_{n}= 0$ b
That the difference between consecutive terms goes to zero does not guarantee convergence. It is necessary but not sufficient. It looks as though you are saying it is necessary, if so that is OK.

CB
 

OhMyMarkov

Member
Mar 5, 2012
83
How can I prove it unbounded? That was my question all along :eek:
 

CaptainBlack

Well-known member
Jan 26, 2012
890
How can I prove it unbounded? That was my question all along :eek:
Look at my first post in this thread (now that the obvious typo has been corrected), it shows that there is a \(c>1\) such that \(x_{n_0+k}>c^k x_{n_0}\) ...

CB
 

OhMyMarkov

Member
Mar 5, 2012
83
Thank you Captain Black, I finally got it now, I'll continue it perhaps someone encounters the problem and comes looking for the solution:

we have $x_{n_0 +k}>c^k x_{n_0}$. But $n = n_0 + k$, we get:
$x_n>c^n\cdot (x_{n_0}c^{-n_0})=\alpha c^n$.

We know $c^n$ converges to $\infty$ because $c=|c|>1$, therefore, $x_n$ converges to $\infty$ as well.
 
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