for all , that means
for all .
I assume since this is a sequence that only takes on postitive integer values, so that would mean is positive and therefore exponentiating would keep the inequality intact.
So yes, the sequence converges.
not necessarily implies that it converges
True, I misinterpreted the question, sorry. All I proved was that it was bounded.
According to Wolfram,
so yes, the sequence converges.
Thanks a lot Sir!
what is the url at Wolfram?
... and because ...
... is ...
is not obvious. there are integer that make very close to hence is very big.
Do we have ?
Reply to Elim at Post # 5:
Prove It is most likely referring to WolframAlpha. It's basically a way to enter single Mathematica commands. At least, that's what I use it for. Quite handy.
... means that for any it exists an integer so that is...
Now if we measure angles in degree [for example] instead of in radians is and the sequence has no limit. If the angles are in radians then never vanishes and in my opinion the (1) is 'probably true', even if for 'very small' the (2) is satisfied for 'enormous' values of . In any case a rigorous prove has to be supplied and that is not a trivial job (Thinking)...