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.
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.
Clearly it diverges so is not Cauchy but
However this is not the sequence Markeur needs because it is not bounded.
Black idea is to define
Let if n is odd or 0 if n is even
for all n thus , it is bounded but divergent.
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.