Does converge?

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- Aug 27th 2010, 05:57 PMelimDetermine whether a sequence converge
Does converge?

- Aug 27th 2010, 07:13 PMProve It
Remember that

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

.

So yes, the sequence converges. - Aug 27th 2010, 07:23 PMelim
not necessarily implies that it converges

- Aug 27th 2010, 07:28 PMProve It
True, I misinterpreted the question, sorry. All I proved was that it was bounded.

According to Wolfram,

so yes, the sequence converges. - Aug 27th 2010, 08:29 PMelim
Thanks a lot Sir!

what is the url at Wolfram? - Aug 28th 2010, 04:39 AMchisigma
Is...

(1)

... and because ...

(2)

... is ...

(3)

Kind regards

- Aug 29th 2010, 09:37 PMelim

is not obvious. there are integer that make very close to hence is very big. - Aug 30th 2010, 01:33 AMqspeechc
Do we have ?

- Aug 30th 2010, 02:55 AMAckbeet
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. - Aug 30th 2010, 05:33 AMelim
- Aug 30th 2010, 05:35 AMelim
- Aug 30th 2010, 10:05 AMchisigma
The consideration of elim is quite correct. By definition say that...

(1)

... means that for any it exists an integer so that is...

(2)

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)...

Kind regards

- Aug 30th 2010, 11:17 AMqspeechc
- Aug 30th 2010, 12:23 PMDefunkt
- Aug 31st 2010, 08:27 AMqspeechc