# Math Help - Limits

1. ## Limits

I really need help with this one. Is there anyone who can get me started? thanks

2. Originally Posted by Sasbe
I really need help with this one. Is there anyone who can get me started? thanks
Well, it's been a few weeks since I've done limits, but if $n$ goes to infinite, then you're basically having every $n$ variable go off to 0, so if you substitute 0 in, you're left with $lim a(0)+1-a(0)/0=0$ and $1/0=0$ so that should be your proof.

3. Originally Posted by emttim84
Well, it's been a few weeks since I've done limits, but if $n$ goes to infinite, then you're basically having every $n$ variable go off to 0, so if you substitute 0 in, you're left with $lim a(0)+1-a(0)/0=0$ and $1/0=0$ so that should be your proof.
I don't get it. It's not possible to divide by 0, right?

4. Originally Posted by Sasbe
I really need help with this one. Is there anyone who can get me started? thanks
i'll try to prove it like this.. since $\lim_{n\rightarrow \infty } \, a_{n+2} - a_n = 0$, then $a_{n+2} - a_n$ is bounded, say by M

$\implies |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M \implies |a_{n+1} - a_n| < M$ (why?) $\implies -M < a_{n+1} - a_n < M$

so when we multiply $\frac{1}{n}$ to all, we have $\frac{-M}{n} < \frac{a_{n+1} - a_n}{n} < \frac{M}{n}$ and if $n \rightarrow \infty$, we have, $0 < \frac{a_{n+2} - a_n}{n} < 0$ and by squeeze theorem, we get $\lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0$. QED

5. Originally Posted by kalagota
i'll try to prove it like this.. since $\lim_{n\rightarrow \infty } \, a_{n+2} - a_n = 0$, then $a_{n+2} - a_n$ is bounded, say by M

$\implies |a_{n+2} - a_n| \leq M \implies -M \leq a_{n+2} - a_n \leq M$

so when we multiply $\frac{1}{n}$ to all, we have $\frac{-M}{n} \leq \frac{a_{n+2} - a_n}{n} \leq \frac{M}{n}$ and if $n \rightarrow \infty$, we have, $0 \leq \frac{a_{n+2} - a_n}{n} \leq 0$ and by squeeze theorem, we get $\lim_{n \rightarrow \infty} \frac{a_{n+2} - a_n}{n} = 0$. QED
Thats right, but did you notice that it was suppose to be
$\lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0$
$\lim_{n \rightarrow \infty} \frac{a_{n+2} - a_n}{n} = 0$
?

6. Originally Posted by Sasbe
Thats right, but did you notice that it was suppose to be
$\lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0$
$\lim_{n \rightarrow \infty} \frac{a_{n+2} - a_n}{n} = 0$
?

In that case you can write,
$a_{n+2}-a_n = (a_{n+2}-a_{n+1})+(a_{n+1}-a_n)$ since $\lim a_{n+1}-a_n = \lim a_{n+2}-a_{n+1}=0$ apply kologata's argument.

7. Originally Posted by ThePerfectHacker
In that case you can write,
$a_{n+2}-a_n = (a_{n+2}-a_{n+1})+(a_{n+1}-a_n)$ since $\lim a_{n+1}-a_n = \lim a_{n+2}-a_{n+1}=0$ apply kologata's argument.
It may indeed matter. Take the example $a_n = \left( { - 1} \right)^n + \frac{1}{n}$.
It is clear that both n & n+2 are together even or odd.
So $\left( {a_{n + 2} - a_n } \right) \to 0$.

However, $\left( {a_{n + 1} - a_n } \right)$ does not converge to zero.
If n is even we get $\left( {a_{n + 1} - a_n } \right) = \left[ { - 1 + \frac{1}{{n + 1}}} \right] - \left[ {1 + \frac{1}{n}} \right] = - 2 + \frac{1}{{n + 1}} - \frac{1}{n}$.
Whereas if n is odd we get $\left( {a_{n + 1} - a_n } \right) = \left[ {1 + \frac{1}{{n + 1}}} \right] - \left[ { - 1 + \frac{1}{n}} \right] = 2 + \frac{1}{{n + 1}} - \frac{1}{n}$.

The key here is being bounded. That is what needs to be shown for $\left( {a_n } \right)$

8. Originally Posted by Sasbe
Thats right, but did you notice that it was suppose to be
$\lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0$
$\lim_{n \rightarrow \infty} \frac{a_{n+2} - a_n}{n} = 0$
?

oh yeah, didn't notice it.. tnx

i've edited the solution, try to check it..

9. Originally Posted by kalagota
i'll try to prove it like this.. since $\lim_{n\rightarrow \infty } \, a_{n+2} - a_n = 0$, then $a_{n+2} - a_n$ is bounded, say by M

$\implies |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M \implies |a_{n+1} - a_n| < M$ (why?) $\implies -M < a_{n+1} - a_n < M$

so when we multiply $\frac{1}{n}$ to all, we have $\frac{-M}{n} \leq \frac{a_{n+1} - a_n}{n} \leq \frac{M}{n}$ and if $n \rightarrow \infty$, we have, $0 \leq \frac{a_{n+2} - a_n}{n} \leq 0$ and by squeeze theorem, we get $\lim_{n \rightarrow \infty} \frac{a_{n+1} - a_n}{n} = 0$. QED
How do we know this?
$|a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M$

I would understand this if $|a_{n+2} - a_{n+1}| + |a_{n+1} - a_n|$ was less (not greater) than $|a_{n+2} - a_{n+1} + a_{n+1} - a_n|$, but its not.

10. Originally Posted by kalagota
i'll try to prove it like this.. since $\lim_{n\rightarrow \infty } \, a_{n+2} - a_n = 0$, then $a_{n+2} - a_n$ is bounded, say by M
$\implies |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M \implies |a_{n+1} - a_n| < M$
That still does not work!
As I said above, you must show (if you can) that $\left( {a_n } \right)$ is bounded.

11. Originally Posted by Plato
That still does not work!
As I said above, you must show (if you can) that $\left( {a_n } \right)$ is bounded.
it is bounded.. take note that a convergent sequence is bounded.. so if you consider $a_{n+2} - a_n$ as a single sequence (convergent to 0), then you can conclude that it is bounded!

12. Originally Posted by Sasbe
How do we know this?
$|a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M$

I would understand this if $|a_{n+2} - a_{n+1}| + |a_{n+1} - a_n|$ was less (not greater) than $|a_{n+2} - a_{n+1} + a_{n+1} - a_n|$, but its not.
yes, it could be equal.. but not on the last parts that i've already edited..
try to group the terms: $|(a_{n+2} - a_{n+1}) + (a_{n+1} - a_n)|$, then use triangle inequality...

13. Originally Posted by kalagota
it is bounded.. take note that a convergent sequence is bounded.. so if you consider $a_{n+2} - a_n$ as a single sequence (convergent to 0), then you can conclude that it is bounded!

The whole point is that $\left( {a_n } \right)$ may not be a covergent sequence.

Take a close look at the example that I gave above.
In that example the $\left( {a_n } \right)$ sequence does not converge. Now it is bounded.

So from the given, you may not conclude that the sequence converges because there is a counterexample.
How do you prove that it is bounded?

14. Originally Posted by Plato
The whole point is that $\left( {a_n } \right)$ may not be a covergent sequence.
Take a close look at the example that I gave above.
In that example the $\left( {a_n } \right)$ sequence does not converge. Now it is bounded.

So from the given, you may not conclude that the sequence converges because there is a counterexample.
How do you prove that it is bounded?
Theorem: A convergent sequence is bounded.. (do i need to prove this?)

the given was $\lim_{n \rightarrow \infty} a_{n+2} - a_n$ and not ${a_n }$ alone..
you should not put your interest with $a_n$ since what we want is the sequence $a_{n+1} - a_n$ and that sequence is different from $a_n$!

if you find any "flaws" with my "proof" while i had used only the given and the definition and some theorems, point it out.. well, i would try to consult this with my professor tomorrow..

15. Originally Posted by kalagota
if you find any "flaws" with my "proof" while i had used only the given and the definition and some theorems, point it out..
OK. Here is the major flaw.
$\color{red} |a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M \implies |a_{n+1} - a_n| < M$
There is absolutely no way that you can claim that. It is a false claim.
Originally Posted by Sasbe
How do we know this?
$|a_{n+2} - a_{n+1} + a_{n+1} - a_n| \leq |a_{n+2} - a_{n+1}| + |a_{n+1} - a_n| \leq M$
I would understand this if $|a_{n+2} - a_{n+1}| + |a_{n+1} - a_n|$ was less (not greater) than $|a_{n+2} - a_{n+1} + a_{n+1} - a_n|$, but its not.