# Convergence of sequences proof

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

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?

Homework Helper
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.)

synkk
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?

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

synkk
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.)

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

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?

synkk
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?
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,

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

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

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

Last edited:
synkk
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.

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|##?

Homework Helper
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|##?

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.

synkk
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.

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

Homework Helper
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

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}##.

synkk
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}##.

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