1. Limit of a Sequence

Is there a sequence (a sub n) which is bounded and is not convergent such that a sub n+1 - a sub n tens to 0 as n tends to infinity?

Thanks.

2. 1,0,1,0,1,0,...

3. Originally Posted by Black
1,0,1,0,1,0,...
You mean (a sub n)=(1,0,1,0,1,0,...) and (a sub n+1)=(1,0,1,0,1,0,...)?

4. Originally Posted by Markeur
You mean (a sub n)=(1,0,1,0,1,0,...) and (a sub n+1)=(1,0,1,0,1,0,...)?
That doesn't even make sense. If (a_n) = (1,0,1,0,1,0,...) how could (a_n+1) equal the very same thing?

Also, lim (a_n) - lim (a_n+1) is not the same as lim(a_n - a_n+1).

Edit: That wasn't right, because it's minus.

Well, if the lim (a_n - a_n+1) = 0, then the series is Cauchy, and therefore convergent. So the answer is no.

5. Originally Posted by davismj
That doesn't even make sense. If (a_n) = (1,0,1,0,1,0,...) how could (a_n+1) equal the very same thing?

Also, lim (a_n) - lim (a_n+1) is not the same as lim(a_n - a_n+1).

Edit: That wasn't right, because it's minus.

Well, if the lim (a_n - a_n+1) = 0, then the series is Cauchy, and therefore convergent. So the answer is no.
Actually,the question that I posed was in Cambridge tripos 2006,Q11.
http://www.math.cam.ac.uk/teaching/p.../PaperIA_1.pdf

Since it was asked in the question,there must be some sequences that have this property.Just that I still can't think of any.

6. Originally Posted by davismj
That doesn't even make sense. If (a_n) = (1,0,1,0,1,0,...) how could (a_n+1) equal the very same thing?

Also, lim (a_n) - lim (a_n+1) is not the same as lim(a_n - a_n+1).

Edit: That wasn't right, because it's minus.

Well, if the lim (a_n - a_n+1) = 0, then the series is Cauchy, and therefore convergent. So the answer is no.

That is not right. A cauchy sequence does have that property but being cauchy is so much stronger.

For example:

$\displaystyle x_n=\sum_{1}^{n} \frac{1}{i}$

Clearly it diverges so is not Cauchy but $\displaystyle |x_{n+1}-x_n|=\frac{1}{n+1} \to 0$

However this is not the sequence Markeur needs because it is not bounded.
Black idea is to define

Let $\displaystyle x_n= 1$ if n is odd or 0 if n is even

then
$\displaystyle x_{n+1}-x_n= 0$ for all n thus $\displaystyle x_{n+1}-x_n \longrightarrow 0$ , it is bounded but divergent.

7. Actually, now that I looked at the link, my example will fail. I think the sequence

$\displaystyle a_n=\text{sin}\left(\sum_{k=1}^n\frac{1}{k}\right)$

will work.

8. Why would it fail? I cant see the reason

9. Because, if you have the sequence 1,0,1,0,... , then $\displaystyle |a_{n+1}-a_n|=1, \, \forall n \in \mathbb{N}$.

10. Ouch LOL

you are right i totally missed that

11. Originally Posted by mabruka
That is not right. A cauchy sequence does have that property but being cauchy is so much stronger.

For example:

$\displaystyle x_n=\sum_{1}^{n} \frac{1}{i}$

Clearly it diverges so is not Cauchy but $\displaystyle |x_{n+1}-x_n|=\frac{1}{n+1} \to 0$

However this is not the sequence Markeur needs because it is not bounded.
Black idea is to define

Let $\displaystyle x_n= 1$ if n is odd or 0 if n is even

then
$\displaystyle x_{n+1}-x_n= 0$ for all n thus $\displaystyle x_{n+1}-x_n \longrightarrow 0$ , it is bounded but divergent.
That's true, about what I said. My mistake. About what you said. (1,0,1,0,...) - (0,1,0,1,...) is (1,-1,1,-1,...), which doesn't converge, let alone to zero.

Regarding black's other idea...

I don't think so. Any way you split up sin(x), the n+1th term will never come arbitrarily close to the nth term without being convergent.

In general, (a_n+1) - (a_n) would require the limit of (a_n+1) to converge to the limit of (a_n) and therefore would require (a_n) to have a limit. Since this is clearly not the case, the only other option is that (a_n+1) = (a_n) for every n, which (since this is pretty much induction) means that the sequence is constant. Here again, the sequence converges.

12. 3, 3.1, 3.14, 3.141, 3.1415... is a bounded sequence, where $\displaystyle (a_{n+1} - a_n) \rightarrow 0$ but isn't convergent in $\displaystyle Q$

But I guess you're talking about sequences in R?

13. Originally Posted by JG89
3, 3.1, 3.14, 3.141, 3.1415... is a bounded sequence, where $\displaystyle (a_{n+1} - a_n) \rightarrow 0$ but isn't convergent in $\displaystyle Q$

But I guess you're talking about sequences in R?
That's a good example, but I think if he was looking for something like that he'd say "Cauchy but not convergent".

14. What is wrong with $\displaystyle a_n=\sin\left(\sum_{k=1}^n\frac{1}{k}\right)$ ?

We have it is not convergent.
It is bounded.
And since sin is continuous, for every $\displaystyle \epsilon > 0$ there is $\displaystyle \delta > 0$ such that

if $\displaystyle \frac{1}{n+1}=\left | \sum_{k=1}^{n+1}\frac{1}{k} -\sum_{k=1}^n\frac{1}{k} \right | < \delta$ then

$\displaystyle \left | \sin\left(\sum_{k=1}^{n+1}\frac{1}{k}\right) -\sin\left(\sum_{k=1}^n\frac{1}{k}\right) \right | < \epsilon$

So for every $\displaystyle \epsilon$ there is an $\displaystyle N \in \mathbb N$ such that if $\displaystyle n+1>N$, then

$\displaystyle \frac{1}{n+1} < \delta$ and

$\displaystyle |a_{n+1}-a_n|=\left | \sin\left(\sum_{k=1}^{n+1}\frac{1}{k}\right) -\sin\left(\sum_{k=1}^n\frac{1}{k}\right) \right | < \epsilon$

15. I wonder whether there is a sequence $\displaystyle x_n$ such that $\displaystyle \displaystyle\lim_{n\to\infty}x_n^{2} = L$ and $\displaystyle x_{n+1}-x_n \longrightarrow 0$ but $\displaystyle x_n$ diverges?

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