# Homework Help: Convergence of sequences proof

1. Oct 19, 2013

### synkk

Given a sequence $<x_n>$, let $<x_{n+1}>$ denote the sequence whose nth term for each $n \in \mathbb{N}$ is $x_{n+1}$. Show that if $<x_n>$ converges then $< x_{n+1}$ converges and they have the same limit.

my attempt thus far

given $\epsilon > 0$ $\exists N \in \mathbb{R}$ s.t. n > N implies $|x_n - l | < \epsilon$ where l is the limit of $x_n$.

now we need to prove that for given $\epsilon > 0$ $\exists N_1 \in \mathbb{R}$ s.t. $n > N_1$ implies $|x_{n+1} - m | < \epsilon$ where m is the limit of $x_{n+1}$ and m = l

heres what I have thus far

consider $|x_n - l |$ :

$|x_n - l | = |x_n - l + x_{n+1} - x_{n+1} | = |(x_n - x_{n+1}) + (x_{n+1} - l ) | \leq |x_n - x_{n+1} | + |x_{n+1} - l| \leq |x - x_{n+1} | + |x_{n+1} - l | \leq |x_{n+1} - x | + |x_{n+1} -l |$

any ideas how to proceed? or if I'm even doing it correctly

2. Oct 19, 2013

### ArcanaNoir

Not a bad try. But you want to show $|x_{n+1}-l|\le \epsilon$ so you should start your inequality with that instead of $|x_n-l|$. Also when you break up the absolute value into two pieces, use what you know about the size of $|x_n-l|$ and $|x_n-x_{n+1}|$. I'm not sure why you introduced "x", what is it?

3. Oct 19, 2013

### HallsofIvy

If $\{x_n\}$ converges to L, then, given any $\epsilon>0$, there exist N such that if n> N then $|x_n- L|< \epsilon$. Here, your second sequence, lets call it $\{a_m\}$ has the property that $a_m= x_{n+1}$. Do you see that, for the same $\epsilon$, we can our new "N" to be the old N plus 1?

(This is a different method than ArcanaNoir's. Both will work.)

4. Oct 19, 2013

### synkk

The x is the first term of the sequence

if I do x_n - x_(n+1) = x_1, x_2, x_3 ... x_n - (x_2, x_3, x_4... x_(n+1))

how do I evaluate $|x_n-x_{n+1}|$? I have no idea of it's size

5. Oct 19, 2013

### synkk

No, I don't see that - I mean it kind of makes sense, but I'm not all that convinced.

6. Oct 19, 2013

### ArcanaNoir

Have you learned about Cauchy sequences?

At any rate, you know $|x_n-l|<\epsilon$ for all n>N. Well if n>N, isn't n+1 > N?

7. Oct 19, 2013

### synkk

I have not learned about Cauchy sequences, all we have done is the definition of convergence and this is a question on my problem sheet.

I agree with the highlighted part, but I do not see how it helps.

thanks,

8. Oct 19, 2013

### Dick

Maybe I just don't get what you don't get, but if

doesn't that mean $|x_{n+1}-l|<\epsilon$? It's really simpler than what you think.

Last edited: Oct 19, 2013
9. Oct 20, 2013

### synkk

Yeah that makes sense actually,

but for a different proof:

I got from here $|x_{n+1} - l| \leq |x_{n+1} - x_n| + |x_n - l | < |x_{n+1} - x_n| + \epsilon$ how do I evaluate $|x_{n+1} - x_n|$?

10. Oct 20, 2013

### Dick

That's really not a very good way to go. Because about the only thing you can say about $|x_{n+1} - x_n|$ is that since $|x_n-l|<\epsilon$ AND $|x_{n+1} - l| < \epsilon$ then $|x_{n+1} - x_n| < 2\epsilon$, but you need the bound on $|x_{n+1} - l|$ to say that.

11. Oct 20, 2013

### synkk

hm I see, but I just cannot seem to construct a proof using that if x_n is bounded then there n>N for some N then n+1 > N also

12. Oct 20, 2013

### Dick

I don't understand why you don't see it. If n>N then n+1>N. It's automatic. The same N that works for the sequence $a_n$ works for the sequence $b_n$ where $b_n=a_{n+1}$.

13. Oct 20, 2013

### synkk

So that's all I'd have to state for the proof?

14. Oct 20, 2013

### Dick

Yes, a proof could be that simple. Write it in a way that you understand it.

15. Oct 20, 2013

### synkk

Thank you very much Dick, and everyone else who helped :)!