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[Mindmaker]

I ran out of books today (well at least readable ones) and I've got a very long and boring day ahead of me.
Not quite sure what I should reread or finish reading.

I could reread "The Wild Gods", a readable renarration, of the old norse edda.
The legends of the norse gods are entertaining, highly educational and fascinating.

Or I can finish reading one of the discworld science novels I never finished reading.

[Armok]

People are being mean to Vectors brain. D:

[KaminaSquirtle]

a₀=2
a_n=((n+3)/(2n))*a_n-1+n/((n+1)*a_n-1), for all integers n>=1
Is the sequence {a_n} convergent?  If so, what is the limit?
If you do solve this, Vector, please don't tell me the answer; the professor is giving out a prize to first student to give him a well written answer, and I want it!

Huh.  Very prettily made.  Thank you, I'll be mulling that one over for a couple days.

EDIT: HAH!  I think I got half of it, perhaps, if I can build an epsilon of the right size.

You're welcome.  Though if you're having trouble figuring it out, it makes me doubt my ability to solve it.  I mean, recursively defined sequences were just introduced this week.

But it's not like I'm strapped for time, very few students have ever solved it.

[Mindmaker]

I'd be happy if I had 3 classes to pick from, which were samey and can branch off into 2 "Prestige"-classes (like Kotor2) that didn't do much, or when the choice doesn't matter (like Planescape: Torment), or when there are a lot of of choices, but again they don't really matter (like Gothic 2).

Make me think and it substracts from the enjoyment I get out of a game.

Fixed for you.

That's just mean D:

I enjoy a game that makes me think.
Sadly I'm a person who tends to overthink.
That's what actually gives me a hard time to make any decision at all.

[Vector]

You're welcome.  Though if you're having trouble figuring it out, it makes me doubt my ability to solve it.  I mean, recursively defined sequences were just introduced this week.

My problem, basically, is that I have stupidly good intuition but a bad memory for whatever tools I've made.  In this case, it looks like a Cauchy sequence to me, but I'm not quite sure what to do with it next.  Well, I do have ideas, but I think the right answer for now is a little sleep.

[Bauglir]

-snip-

[KaminaSquirtle]

You're welcome.  Though if you're having trouble figuring it out, it makes me doubt my ability to solve it.  I mean, recursively defined sequences were just introduced this week.

My problem, basically, is that I have stupidly good intuition but a bad memory for whatever tools I've made.  In this case, it looks like a Cauchy sequence to me, but I'm not quite sure what to do with it next.  Well, I do have ideas, but I think the right answer for now is a little sleep.

Well, this is the last problem before the chapter on Cauchy sequences, so I don't think you'll need to use anything involving them.  I'd go for proving it monotonic and bounded, and going from there.  That's really the only way I know to solve recursive sequences.  However, I still haven't begun the problem yet, so I don't know if that'll be crazy hard or not.  Actually, I haven't delt with sequences that are recursive and have factors of n in them yet, so I'll have a fun time with this one.  I'll work my way through some other problems before I try to tackle this one.

[Aqizzar]

Just came back from the optometrist.  Same guy I've been going to for years, and he got old as dirt in the meantime.  Apparently my astigmatism is getting worse.

Also ridiculously behind on all the Spanish stuff I'm supposed to be doing, and writing a letter to Strife.  What am I doing instead?  MSPA art dump.  Eh, it's fun and quick so who cares.

[Vector]

Don't think it's monotonic, man.  But maybe.

[KaminaSquirtle]

I'm inclined to think it is, as the chapter it's in is called "monotonic sequences", but it might not be.  It is supposed to be hard, after all.  But as I said, I haven't tried it yet, so I don't know.  I think I'll go mess around with it right now, in fact, see if that gives me some insight.

[Virex]

a0=2
a_n=((n+3)/(2n))*a_n-1+n/((n+1)*a_n-1), for all integers n>=1



My guts tell me that if you take the limit of this function for n to +infinity it's non-a_n factors converge to zero. What I can't tell you is if a grows fast enough to overcome that or not though. scrap that I'm going too fast

[Vector]

a0=2
a_n=((n+3)/(2n))*a_n-1+n/((n+1)*a_n-1), for all integers n>=1



My guts tell me that if you take the limit of this function for n to +infinity it's non-a_n factors converge to zero. What I can't tell you is if a grows fast enough to overcome that or not though.

Which means that the second term explodes.

As I said, I think it converges.

[Virex]

Spoiler: in case someone wants to solve it for himself, don't look here (click to show/hide)

I went to fast, first term converges to a_n-1/2 for n -> inf, because (n+3)/(2n) = 1/2 + 3/(2n) for n != 0
Second term converges to 1 for non-infinite values of a_n (lim n-> inf n/(n+c) = 1 if c << n)
This means that for sufficiently large values of n, the function converges to a_n = a_n-1/2 + 1, meaning a_n will rise unbound.
I don't know for sure what happens if at some point a_n-1>>n, because that would mean that the second term converges to 0, causing the function to converge to a_n = a_n-1/2, reducing a_n again.

[Vector]

Fast question: have you solved it completely?

[Virex]

Not yet, as I'm not entirely sure what happens. I think I know what happens, but I can't formulate it properly.

Spoiler (click to show/hide)

I suspect the function doesn't converge, because if a_n goes to infinity, the function reduces to the second form, which forces a_n down again and the cycle repeats. I can't prove that it doesn't converge to a constant though