1. ## Sequence Proof

Prove that if x_n --> infinity, then the sequence given by

x_n / x_n+1 is convergent.

I am supposed to use the definition of divergence, so I know there is a number n for which all x_n > M for any M chosen.

2. ## Re: Sequence Proof

Originally Posted by veronicak5678
Prove that if x_n --> infinity, then the sequence given by

x_n / x_n+1 is convergent.

I am supposed to use the definition of divergence, so I know there is a number n for which all x_n > M for any M chosen. i think you are using an incorrect definition of divergence
to support what i have written in red:
define $\displaystyle x_n$ as $\displaystyle x_{2k}=0, x_{2k+1}=1 \, \forall k \in \mathbb{Z}^+$.
then $\displaystyle x_n$ is divergent since its not convergent. But if we take $\displaystyle M=5$(say) we can't find any $\displaystyle n$ which would satisfy $\displaystyle x_n>M$.

Also can you please explicitly write what definition of $\displaystyle x_n \rightarrow \infty$ you are using??

3. ## Re: Sequence Proof

Originally Posted by veronicak5678
Prove that if x_n --> infinity, then the sequence given by

x_n / x_n+1 is convergent.
Consider the sequence $\displaystyle x_n = \left\{ \begin{gathered} n,\text{ n odd} \hfill \\ 2^n ,\text{ n even} \hfill \\ \end{gathered} \right.$.

Is it true that $\displaystyle \left( {x_n } \right) \to \infty ~?$

What can you say about $\displaystyle \frac{{x_n }}{{x_{n + 1} }}~?$

4. ## Re: Sequence Proof

Thanks for the answers. The definition of a limit at infinity from my book is as follows:

sn → ∞ as n → ∞
provided that for every number M there is an integer N so that
sn ≥ M
whenever n ≥ N.

5. ## Re: Sequence Proof

Originally Posted by veronicak5678
The definition of a limit at infinity from my book is as follows:
sn → ∞ as n → ∞
provided that for every number M there is an integer N so that
sn ≥ M whenever n ≥ N.
That definition is standard. Now apply it to the example I gave you.
It will show that $\displaystyle (x_n)\to\infty$.

BUT does $\displaystyle \frac{{x_n }}{{x_{n + 1} }}$ converge as claimed.

6. ## Re: Sequence Proof

Not sure I follow...
We could use the same definition to show that
x_n-> inf implies x_n+1-> inf, but how does that prove that x_n / x_n+1 converges?

7. ## Re: Sequence Proof

Originally Posted by veronicak5678
Not sure I follow...
We could use the same definition to show that
x_n-> inf implies x_n+1-> inf, but how does that prove that x_n / x_n+1 converges?
Which of these two do you mean:
$\displaystyle A~\left( {\frac{{x_n }}{{x_{n + 1} }}} \right)\text{ or }B~\left( {\frac{{x_n }}{{x_n + 1}}} \right)$

If it is A then the statement is false.
If it is B then the proof is trivial.

8. ## Re: Sequence Proof

Sorry for being unclear. I mean B.

9. ## Re: Sequence Proof

Originally Posted by veronicak5678
Sorry for being unclear. I mean B.
If $\displaystyle (x_n)\to\infty$ then $\displaystyle \left( {\frac{1}{{x_n+1 }}} \right) \to 0$.

Note that $\displaystyle \left( {\frac{{x_n }}{{x_n + 1}}} \right) = \left( {1 - \frac{1}{{x_n + 1}}} \right)$

10. ## Re: Sequence Proof

I understand that, but how can I prove this using just the definition of divergence, and no theorems about limits?

11. ## Re: Sequence Proof

Originally Posted by veronicak5678
I understand that, but how can I prove this using just the definition of divergence, and no theorems about limits?
If $\displaystyle \varepsilon > 0$ use the divergence definition to make $\displaystyle x_n + 1 > \frac{1}{{1 + \varepsilon }}$.

Having to do it this is just busy work.
So that is as far as I am willing to take it.

12. ## Re: Sequence Proof

OK. Thanks for helping me.