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Problem 17 section 3.2 Bartle

issacnewton

Member
Jan 30, 2012
61
here is the problem from Bartle's book
Let \( X=(x_n) \) be a sequence of positive real numbers such that \( \mbox{lim }(x_{n+1}/x_n) = L > 1 \). Show that \( X \) is not bounded sequence
and hence is not convergent.

I am using negation of the goal. So I assumed that the sequence is bounded. In the limit definition, by using \( \varepsilon =L-1 \) I could show that

\[ \exists n_1 \in \mathbb{N}\;\forall \;n\geqslant n_1 \; (x_n < x_{n+1}) \]

Can people give some more hints ?
 

Alexmahone

Active member
Jan 26, 2012
268
It's easy to show that $X$ is not convergent.

Assume, for the sake of argument, that $X$ converges to $l$.
$\displaystyle\lim\left(\frac{x_{n+1}}{x_n}\right)=\frac{l}{l}=1$, a contradiction.
 
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Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,488
Alternatively, you need to know that eventually we have not just $x_{n+1}/x_n>1$, but $x_{n+1}/x_n>a$ for some a > 1.
 

issacnewton

Member
Jan 30, 2012
61
Alex , how does that follow ? Which theorem you are using here ?
 

Alexmahone

Active member
Jan 26, 2012
268
Alex , how does that follow ? Which theorem you are using here ?
$\displaystyle \lim\left(\frac{x_{n+1}}{x_n}\right)=\frac{\lim\ x_{n+1}}{\lim\ x_n}$ (Limit of a quotient equals quotient of limits.)
 

issacnewton

Member
Jan 30, 2012
61
Alternatively, you need to know that eventually we have not just $x_{n+1}/x_n>1$, but $x_{n+1}/x_n>a$ for some a > 1.
Since \( L-1 > 0 \) , we can choose some rational ( or irrational) between L-1 and 0. We can choose this number as our \( \varepsilon \). So \( 0< \varepsilon < L-1 \). Using the information given, we can choose some \( n_1 \in \mathbb{N} \) such that for all \( n\geqslant n_1 \) we have

\[ L-\varepsilon < \frac{x_{n+1}}{x_n} < L+\varepsilon \]

Now define \( a= L-\varepsilon \). Since \( \varepsilon < L-1 \) , we have \( 1 < L- \varepsilon \) , so \( \therefore a > 1 \). Hence
\( a < \frac{x_{n+1}}{x_n} \) for some \( a > 1 \) as you suggested. So how does it help ?
 

Alexmahone

Active member
Jan 26, 2012
268
Since \( L-1 > 0 \) , we can choose some rational ( or irrational) between L-1 and 0. We can choose this number as our \( \varepsilon \). So \( 0< \varepsilon < L-1 \). Using the information given, we can choose some \( n_1 \in \mathbb{N} \) such that for all \( n\geqslant n_1 \) we have

\[ L-\varepsilon < \frac{x_{n+1}}{x_n} < L+\varepsilon \]

Now define \( a= L-\varepsilon \). Since \( \varepsilon < L-1 \) , we have \( 1 < L- \varepsilon \) , so \( \therefore a > 1 \). Hence
\( a < \frac{x_{n+1}}{x_n} \) for some \( a > 1 \) as you suggested. So how does it help ?
$\displaystyle x_{n+1}>ax_n$ for $\displaystyle n\ge n_1$

$\displaystyle x_{n_1+k}>a^kx_{n_1}$ for $k>0$ (Use induction)

Since $\displaystyle a>1$, $\displaystyle a^k$ can be made arbitrarily large by choosing a large enough $\displaystyle k$. Can you finish off?
 
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issacnewton

Member
Jan 30, 2012
61
Ah, I see \( (x_{n+1}) \) is 1 tail of sequence \( x_n \) and by the related theorem both converge to the same number. So ok, that proves that
\( X \) doesn't converge to any number. Now what about the boundedness, since even divergent sequences are bounded ...

---------- Post added at 11:41 PM ---------- Previous post was at 11:25 PM ----------

Alex, yes I will work on it. Just a quick question. Assuming I prove that \( a^k \) can be made arbitrarily large. That is, given any \( M>0 \), we can get \( k \) such that \( a^k > M \). That means \( a^k x_{n_1} > M x_{n_1} \) since \( x_{n_1} > 0 \). Now how does it help to show that
\( x_{n_1+k} > M \) ?
 

Alexmahone

Active member
Jan 26, 2012
268
Alex, yes I will work on it. Just a quick question. Assuming I prove that \( a^k \) can be made arbitrarily large. That is, given any \( M>0 \), we can get \( k \) such that \( a^k > M \). That means \( a^k x_{n_1} > M x_{n_1} \) since \( x_{n_1} > 0 \). Now how does it help to show that
\( x_{n_1+k} > M \) ?
Given any $M>0$,

$x_n>M$ for $n>n_1$

if $\displaystyle x_{n_1+k}>M$

if $\displaystyle a^kx_{n_1}>M$

if $\displaystyle a^k>\frac{M}{x_{n_1}}$

if $\displaystyle k>\log_a\frac{M}{x_{n_1}}$

if $\displaystyle n>n_1+\log_a\frac{M}{x_{n_1}}$

This completes the proof that $X$ is not bounded.
 
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issacnewton

Member
Jan 30, 2012
61
here is another approach. since \( \forall n\in \mathbb{N}\; x_n > 0 \), define a new sequence. \( t_n = \frac{1}{x_n} \). So \( \frac{t_{n+1}}{t_n} = \frac{x_n}{x_{n+1}} \). Now define another sequence, \( z_n =\frac{t_{n+1}}{t_n} \). So \( z_n=\frac{1}{(\frac{x_{n+1}}{x_n})} \). Now for all n in N, we have \( \frac{x_{n+1}}{x_n} > 0 \). and \( \lim \frac{x_{n+1}}{x_n} = L \neq 0 \). So we can use the limit theorem for the quotient. And we get \( \lim \frac{t_{n+1}}{t_n}= \frac{1}{\lim \frac{x_{n+1}}{x_n} } = \frac{1}{L} < 1 \). Since \( \forall n\in \mathbb{N}\; x_n > 0 \), we have
\( \forall n\in \mathbb{N}\; t_n > 0 \). So with the help of another theorem, it follows that \( \lim (t_n) = 0 \). Now to prove that \( X \) is not bounded, let \( M>0 \) be arbitrary. So \( \frac{1}{M} > 0 \). Since \( \lim (t_n) = 0 \), there exists \( n_1\in \mathbb{N} \) such that , for all \( n\geqslant n_1\), we have \( |t_n| < \frac{1}{M} \). Since \( t_n > 0 \) for all n, we have \( t_n < \frac{1}{M} \). Which means \( \frac{1}{x_n}<\frac{1}{M} \)
\(\Rightarrow x_n > M \). So \( \exists n_1\in \mathbb{N}\; \forall n\geqslant n_1 \; (x_n > M )\). Since M is arbitrary, this proves that \( X\) is not bounded and hence not convergent.