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