# Problem 17 section 3.2 Bartle

#### issacnewton

##### Member
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
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
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
Alex , how does that follow ? Which theorem you are using here ?

#### Alexmahone

##### Active member
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
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
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
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
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
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.