1. ## Need confirmation.

I answered a question but I don't know if my answer is correct so I would appreciate if someone can confirm it. Thank you.
By the way I did some mistake when I wrote it. its an+1-a​n

2. ## Re: Need confirmation.

suppose

$a_n=(-1)^n$

then

$\forall n (a_{n+1}+a_n)=0$

certainly $a_n$ doesn't converge. It is however bounded. I don't think you can play the same trick if {a} is unbounded. I'd have to think about it.

3. ## Re: Need confirmation.

Hmmm, I didn't think about that, I thought if it didn't work I had to come to a contradiction algebrically from the definition of CAUCHY or something .. So in this case I have to give a counter-example ?

4. ## Re: Need confirmation.

Does this mean $a_n$ converges to any point? No, we showed that.
Does this mean $a_n$ is bounded? We haven't shown it but I suspect so. Show this.

5. ## Re: Need confirmation.

Originally Posted by romsek

Does this mean $a_n$ converges to any point? No, we showed that.
Does this mean $a_n$ is bounded? We haven't shown it but I suspect so. Show this.
Looks like I'm wrong.

$a_n=(-1)^n log(n)$

clearly $a_n$ is unbounded.

$a_n+a_{n+1}=(-1)^n log(n) + (-1)^{n+1} log(n+1)$

let n be even

$a_n+a_{n+1}=log(n)-log(n+1)=log(\frac{n}{n+1})\rightarrow log(1)=0$

similar if n is odd

So you can manage to make the sum of adjacent elements converge to 0 even while the sequence itself is unbounded.

6. ## Re: Need confirmation.

Let $b_n = a_{n+1} + a_n$. According to the question, $\lim_{n \to \infty} b_n = 0$. As romsek showed, the convergence of $b_n$ has nothing to do with the convergence of $a_n$. But, we can use $b_n$ to look at properties of $a_n$.

Check out $\lim_{n \to \infty} (b_{n+1} - b_n)$. By limit laws, the limit of the difference is the difference of the limits, so long as both limits exist. Since both do, we know that $\lim_{n \to \infty} (b_{n+1} - b_n) = \lim_{n \to \infty} (a_{n+2} - a_n) = 0$. You can do the same for $\lim_{n \to \infty} (b_{n+2} - b_{n+1}) = \lim_{n \to \infty} (a_{n+3} - a_{n+1}) = 0$. So, if you define sequences $c_n = a_{2n}, d_n = a_{2n+1}$, perhaps you can show that both of these sequences are Cauchy (using the previous results), and hence they both converge. Since convergent sequences are bounded, the original sequence must be bounded, as well.

7. ## Re: Need confirmation.

Hmmmm, maybe we can take the sequence an=n , so we can say that an+1-an converges to 1? but an isn't bounded? Is this possible? i mean i know infinite minus infinite in limits is a problem but the distance between them should be 1 and it shouldn't be a problem right?(I got confused , this is completely wrong

8. ## Re: Need confirmation.

Originally Posted by SlipEternal
... perhaps you can show that both of these sequences are Cauchy (using the previous results), and hence they both converge.
As romsek showed in post #5, you obviously cannot show that $c_n,d_n$ are convergent. Oh well. Disproof by counterexample is always a good lesson!

9. ## Re: Need confirmation.

Why are you looking at the sum? it says the difference between them converges to 0

10. ## Re: Need confirmation.

So when I posted I wrote the sum between them but I corrected it by writing a note on the post, did you guys read it ?

11. ## Re: Need confirmation.

So this is the correct one , although I corrected myself when I submitted the thread, sorry for confusing

12. ## Re: Need confirmation.

say what? This is an entirely different problem.

In this case it's just a Cauchy sequence that clearly converges to a point and is thus bounded.

This proof must be on the web in a dozen places or so. Here.