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[Ochita]

Darn it. I dont understand most of this stuff. Probably because I'm only 15 and working at a highsh B. (English marking system.) Still I could try my hand at helping. I might have a weird way of doing it though.

[dragnar]

I might have a weird way of doing it though.

All I can see while reading this is Fighter messing something up horribly while saying "I'm a helper!"
...Not sure why.

[Ochita]

I might have a weird way of doing it though.

All I can see while reading this is Fighter messing something up horribly while saying "I'm a helper!"
...Not sure why.

Thats.....Comforting.

[MaximumZero]

_{Yeah, I know that I suck at Paint.net, but what do you expect for two minutes?}

[Virex]

Continuing from the "things that made you sad today" thread because we're freaking people out

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

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


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 1/a_n-1 for non-infinite values of a_n (lim n-> inf n/(d*n+c) = 1/d if c << n)
This means that for sufficiently large values of n, the function converges to a_n = a_n-1/2 + 1/a_n-1

I found an error in my original reasoning

Spoiler: continuing my reasoning, almost there but not yet (click to show/hide)

If n -> inf, we get:
a_n = a_n-1/2 + 1/a_n-1 for values of a_n-1<< n and a_n-1/2 for values of a_n-1 >> n. This means that if the function can actually converge a value x exists for which x/2 + 1/x = x, which is
1/2 x² + 1 = x²
1 - 1/2 x² = 0
x = sqrt(2)


We also know that for values of a_n-1 far above sqrt(2), the function shrinks exponentially, while for values of a_n-1 close to 0, the second term of the given function will be largely positive and thus the function will revert to the case of a_n-1 above sqrt(2). Only thing left to do is to prove that it will actually get closer and closer to sqrt(2), so that the limit is actualy defined.

[KaminaSquirtle]

Since I don't know what's under that spoiler, here's the original problem, just in case.

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?

And please keep answers in spoilers, I want to do this one on my own.
Edit:

Spoiler: I think I'm getting somewhere (click to show/hide)

Hmmm, it would seem to be decreasing for all n>=4.
Now if I can just prove it, then I'll have a start at least.
Hopefully I'm at least on the right track...

Edit again:

Spoiler (click to show/hide)

Attempting to prove this by induction seems to have been a mistake.
I think I'll have to take a break from this one for a while, do some easier ones, get a better feel for recursively defined sequences.

[Vector]

Induction definitely a mistake >_>  It's completely the wrong sort of problem for that.

I'm not sure on how to do it either yet, though, so I'll post some meanderings up next time I have data.

[KaminaSquirtle]

I meant proving that it is decreasing for all n>=4 by induction, not the whole thing, that would be silly.  But yeah, that didn't work out.  It might be possible, but it'd be hella ugly.
I've only done very simple recursive sequences, though, so that's why I'm going to go back and do some simpler ones, work my way up.  Maybe enlightenment will come in a future lecture, though I don't have one until Monday.

Just don't make your meanderings too specific please, I don't want to feel like I cheated or anything.

[qwertyuiopas]

Spoiler: I think it might be the solution, depending on how badly I have ignored how math works. (click to show/hide)

infinity - 1 is still infinity, so a_infinity-1 will probably be the same as a_infinity.
After that, algebra brings it to the square root of 2, and the PowerToy calculator on 512-digit precision gives

1/sqrt(2) + sqrt(2)/2 - sqrt(2)
   -3.7337435355230201053445084460834003797620205502706718261256871654238051910489621340849072295529677787238897041770213277037935564375426549458475381341191260877830182961465827513676045397394206930103644635605216383688653644082580389928756266495814879648798849319568560021675957962281933650865185438507995064359744828508382887810224892502880781267864709275527040067654857783869494420459048265844462689025172233812319365260636813344567771302993441521446574068931585902150510582661098100679564894086047283597010732094e-532

and -2.7e532 is far smaller than it's precision (only 512), so I can assume that the resulting number is probably a rounding error.
(Also, I just noticed that 2^1/2/2 = 2^-1/2, so it should equal 0)

Edit: Looked some unicode.

Spoiler: My (probably wrong) reasoning, now with the steps I used. (click to show/hide)

Assumption that probably breaks everything: ∞±1 is still ∞.
a_n=a_n-1*(n+3)/2n + n/a_n-1(n+1)
x->∞: a_∞=a_∞-1*(∞+3)/2∞ + ∞/a_∞-1(∞+1)
  a_∞=a_∞*∞/2∞ + ∞/a_∞∞
  a_∞=a_∞*1/2 + 1/a_∞
  1 = 1/2 + 1/a_∞²  (devision by a_∞)
  1 - 1/2 = 1/a_∞²
  1/2 = 1/a_∞²
  2 = a_∞²
  2^1/2 = a_∞

[Vector]

^ Math butchered irrevocably, sorry.

[KaminaSquirtle]

Every time someone butchers math that badly, a mathematician dies.  DO YOU WANT TO KILL INNOCENT MATHEMATICIANS?!  DO YOU?!

[Vector]

I guess I'd better explain why...

First off, "probably" is not an acceptable method of proof.

Second off, your "algebra" needs to be outlined much more clearly.

Third off, proof by calculator is not a viable method of proof.

Fourth... proof by rounding error is not a viable method of proof.

[qwertyuiopas]

There you go, I found some unicode and edited the steps I have on paper in.

[Vector]

There you go, I found some unicode and edited the steps I have on paper in.

Ah, okay.  Thanks for the revision.

Now the problem is that you're using infinity as a number, rather than a concept.  For example, ∞/∞ is undefined; it doesn't equal 1.  Similar troubles with the use of "probably."

Good try, though--it's an interesting approach (and no, I'm not just saying that to be nice).

[KaminaSquirtle]

YOU DID IT AGAIN, YOU MIGHT KILL MY PROF NEXT, SO PLEASE JUST STOP, THINK OF THE POOR MATHEMATICIANS.
I will point out the most obvious error:
∞/∞ =/= 1, it is an indeterminate form.

"Warning - while you were typing 2 new replies have been posted. You may wish to review your post."

Ninja'd by Vector, who explained things much better than I.