1. Dec 14, 2006

### MathematicalPhysicist

we have a sequence {a_n}, such that for every n natural, a_n>0 and it satisfies:
lim (a_n*a_n+1)=1
prove/disprove:
if {a_n} is bounded then {a_2n} converges.

i havent found any counter example, is this statement true or false, if is false then what's the counter example?

p.s
couldn't you say that this sequence is bassically a sequence of a positive number and its reciprocal?

2. Dec 14, 2006

### MathematicalPhysicist

ofcourse also, a_n=1-1/n satisfies the priliminary criteria, and it's not a number and its reciprocal.
but still i couldn't find a counter example.

3. Dec 15, 2006

### murshid_islam

you could have said that if anan+1 = 1. thats not what is given. but lim (anan+1) = 1. the two are different.

4. Dec 15, 2006

### MathematicalPhysicist

i know that, but do you have a counter example for the statement that if
{a_n} is bounded then {a_2n} converges?

5. Dec 15, 2006

### MathematicalPhysicist

so...???
anyone?!

6. Dec 15, 2006

### kesh

it's a great question. without boundedness 1/1, 2, 1/3, 4, 1/5, 6,... would be a counterexample.

i haven't found a solution, but my hunch is there is no counter example. this means proving it and that means (by above counter example) we have to use the bounded property. this is tricky.

scratch that i think i have a CE

a_n = 2+Sin [Ln[n]], n even

a_n = 1/(2+Sin[Ln[n]]), n odd

idea is anything that oscilates without converging (hence sin), but with the each term getting closer, but not converging (hence log), and its reciprocal. haven't checked it, lol

Last edited: Dec 15, 2006
7. Dec 15, 2006

### MathematicalPhysicist

the whole point of this problem is that it's bounded.
the problem if it's correct then how to prove it, but im not sure that's correct so i asked for counter example.

8. Dec 15, 2006

### kesh

see my edit to my above post

9. Dec 15, 2006

### MathematicalPhysicist

but does lim a_n*a_n+1=1
if n is even then a_n=2+sin(ln(n)) a_n+1=1/(2+sin(ln(n+1))
lim (2+sin(ln(n))/(2+sin(ln(n+1)) does this converges to 1?
ln(n)diverges but sin(ln(n)) it's limit isnt defined.

10. Dec 15, 2006

### kesh

i dunno. it looks like it does, but you need to do the legwork, it's your quiz. it's probably easier to define a similar sequence in a more awkward way that makes working out the product limit easier. the idea is that even terms follow a non convergent path (such as sin wave) but with terms ever closer along such a path, but without converging. odd terms, being the reciprocal of a point between the two even terms, should allow the limit of the product to converge. i'm not sure it's right, but i can't be bothered with pursuing this line tonight
which is what we want

11. Dec 16, 2006

### MathematicalPhysicist

i don't think you example works, cause im not sure it stisfies lim a_n*a_n+1=1.

12. Dec 16, 2006

### kesh

i'm taking that a_n+1 to be the n+1 term of the sequence, not the n term plus 1

like i say i haven't done an epsilon delta proof as the manipulation is messy. i just took it as far as mathematica would go in reasonable time and got it limiting to 1. certainly not a proof, just a good candidate for one

do you get the basic idea of an oscilating a_2n subsequence with terms "getting closer", but not converging?

13. Dec 16, 2006

### kesh

another even subsequence that might do it. 2, 3, 2 1/2, 2, 2 1/3, 2 2/3, 3, 2 3/4, 2 1/2, 2 1/4, 2, 2 1/5, 2 2/5, 2 3/5, 2 4/5, 3, etc

set the odd members appropriately

14. Dec 16, 2006

### kesh

in fact if you set odd terms to just be 1/even terms you get a_n*a_n+1 = 1 or a_n * a_n+1 = (2 + i/p)/(2+(i-1)/p) = (2p + i)/(2p + i-1) which definitely limits to 1. you also have to take care of the "corners" of the oscilation, but i don't see these being a problem

15. Dec 16, 2006

### MathematicalPhysicist

yes i know, but according to your sequence:
when n is even then a_n=2+sin(ln(n))
and n+1 is odd ofcourse so we have (the n+1 term)a_n+1=1/2+sin(ln(n+1))
i dont see how the limit of this multiplication gives one, even by arithematics of limits.

16. Dec 16, 2006

### StatusX

It does work out. Remember ln(n+1)-ln(n)->0 as n->infinity (to see this, just exponentiate it), and since sin is uniformly continuous (which follows from being continuous and periodic), we have that for any e>0 there is a d>0 with |x-y|<d => |sin(x)-sin(y)|<e. Combining these gives the result.

17. Dec 16, 2006

### kesh

you missed some brackets/ that limit doesn't have to give 1. you've forgotten your own question, the product of two succesive (so even and odd) terms has to limit to 1

and StatusX a big thankyou for formalising my intuition, if that's what you did (hungover)

and OP i'm unsubscribing. if you don't get it now you never will

Last edited: Dec 16, 2006
18. Dec 17, 2006

### MathematicalPhysicist

i dont understand stauts, we have something like this:
[2+sin(ln(n))]/[2+sin(ln(n+1)]
we don't have here ln(n+1)-ln(n) outside the sin function, how do you justify that this converges to one, i still dont see it.
perhaps ln(n+1)->ln(n)
and thus sin(ln(n+1))->sin(ln(n)) cause sin is continuous.

ok, i think i get.
perhaps someone knows an example which doesnt use the fact of continuity of functions?

19. Dec 17, 2006

### AKG

$$1, 3,\, 2.5^{-1},\, 2,\, 2.25^{-1},\, 2.5,\, 2.75^{-1},\, 3,\, \left (2\frac{5}{6}\right )^{-1},\, 2\frac{2}{3},\, 2.5^{-1},\, 2\frac{1}{3},\, \left (2\frac{1}{6}\right )^{-1}, 2, 2.125^{-1},\, 2.25,\, \dots$$
The 1 at the beginning serves little purpose except to make (a2n) consist of the terms that oscillate between 3 and 2, and never converge. But the overall sequence is bounded, and at a glance, it appears to satisfy the property $\lim _{n\to \infty}a_na_{n+1} = 1$.