sequence (convergence)

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Jan 10th, 2003, 08:44 PM
bugzpodder

sequence (convergence)

i want to prove if this sequence is convergent or divergent, and if convergent, find its limit.

sqrt(1), sqrt(1+sqrt(2)), sqrt(1+sqrt(2+sqrt(3))), ...

well i guess the first step is to write the general term, but i don't exactly know how to do that. help??
Jan 13th, 2003, 11:37 AM
Zifnab

It's actually not necessary to write it in a general term. You know that sqrt(1) = 1 and that sqrt(1+sqrt(2)) > sqrt(1), the same is true for sqrt(1+sqrt(2+sqrt(3))) (being greater than sqrt(1)) and so on. Hence the remaining statements are all greater than 1. Adding a sequence of n values that are all minimum 1 must go to to infinity for n -> infinity and the sequence is thereby divergent.
Jan 13th, 2003, 12:47 PM
krtxmrtz

Quote:

Originally posted by Zifnab
It's actually not necessary to write it in a general term. You know that sqrt(1) = 1 and that sqrt(1+sqrt(2)) > sqrt(1), the same is true for sqrt(1+sqrt(2+sqrt(3))) (being greater than sqrt(1)) and so on. Hence the remaining statements are all greater than 1. Adding a sequence of n values that are all minimum 1 must go to to infinity for n -> infinity and the sequence is thereby divergent.

Each term is greater than the previous one and larger than one but you're not adding them directly, nor are you adding 1 every time, so your conclusion is not necessarily right. In fact, the series does apparently converge (try writing some code to calculate it to a large n) but I haven't had tiem yet to think of a proof. Hopefully I'll have a good entertainment for this evening... :D
Jan 13th, 2003, 03:17 PM
Zifnab

Quote:

Originally posted by krtxmrtz
Each term is greater than the previous one and larger than one but you're not adding them directly, nor are you adding 1 every time, so your conclusion is not necessarily right. In fact, the series does apparently converge (try writing some code to calculate it to a large n) but I haven't had tiem yet to think of a proof. Hopefully I'll have a good entertainment for this evening... :D

Oops my bad....I took for an infinite series for some reason. Ok, then it's a bit more tricky :D. If no one comes up with a solution tonight I'll have a look at it tomorrow.
Jan 14th, 2003, 03:06 AM
krtxmrtz
I couldn't come up with a proof but the problem has its math-appeal so I'll keep trying evry once in a while. I think it must be very difficult to prove if the general term is not known, maybe you've got to undergo a sudden flash of inspiration, you know, on the lines of Pythagoras' "Eureka" episode.

In the meantime I have verified the convergence by means of this simple code (to calculate the n^th term, place n in a textbox and then place a command button on the form as well):
VB Code:
```
Private Sub Command1_Click()
    n = Trim(Text1.Text)
    a = 0
    For j = n To 1 Step -1
        a = Sqr(a + j)
    Next
    Text1.Text = Str(a)
End Sub
```
This rapidly (n<20) converges to about 1.7579...
Jan 14th, 2003, 03:17 AM
Spooner

Archimedes
Jan 14th, 2003, 03:19 AM
krtxmrtz

Quote:

Originally posted by Spooner
Archimedes

Right, thanks, I'm still asleep this morning.
Jan 14th, 2003, 12:15 PM
bugzpodder

true that value was what i got also but i am looking for a mathematical proof of course.

btw it is possible to express the general term but it isn't very nice
Jan 14th, 2003, 01:10 PM
krtxmrtz

If you know the general term post it, it could help for trying to prove the convergence. I'd like to derive it myself but seems hard and it can wait till I go into the bathtub like Archimedes... :)
Jan 14th, 2003, 02:02 PM
Zifnab

You're right that it's convergent and it looks to converge to the value of 2. I've made an incomplete proof which at least should give an idea of a possible way to prove it:

Take the following series:

a(n) = sqrt(2), sqrt(2 + sqrt(2)), sqrt(2+sqrt(2+sqrt(2)))...
it can be written as:

a(1) = sqrt(2) , a(n+1) = sqrt(2 + a(n))

Since a(1) < sqrt(2) and a(n) < 2
=> a(n+1) = sqrt(2 + a(n)) < sqrt(2 +2) = 2

yet a(n+1) = sqrt(2+a(n)) > sqrt(a(n) + a(n)) , (since a(n) < 2)

= sqrt(2*a(n)) > sqrt(a(n)^2) , (since a(n) < 2)

= a(n)

a(n+1) > a(n) , a(n) < 2, and a(n+1) < 2 so
it's convergent to the value 2.

Then there's the series you proposed, let's call it

b(n) = sqrt(1), sqrt(1+sqrt(2)), sqrt(1 + sqrt(2+sqrt(3))

Then you can see that a(n) > b(n) since if we define
b(n) = sqrt(1 + c(n)) then (I'm not going to
write this out - hopefully it's clear) c(n) < 2. Therefore
b(n) < sqrt(1+2) < a(n).

So now we have a series that's greater which converges to 2.
Now the next thing is to find a series (for example call it d(n)) which also converges to 2 and where b(n) > d(n). If this is can be done then b(n) must also converge to 2. Don't know if it's possible to find such as series, but I believe it is (at least if this series does converge to 2). Anyway those were my 2 cents, interesting prob.
Jan 14th, 2003, 04:41 PM
bugzpodder

the general term is a_n=sqrt(1+sqrt(2+...+sqrt(n))))) ;)
Jan 15th, 2003, 05:50 AM
krtxmrtz

Quote:

Originally posted by bugzpodder
the general term is a_n=sqrt(1+sqrt(2+...+sqrt(n))))) ;)

Well, yes, this we already knew. What I meant was, is it possible to express the general term in terms of the previous one, i.e. a_n = f(a_n-1) ? Or, in such a fashion that you don't need to write the "..." ?
Jan 15th, 2003, 06:23 AM
krtxmrtz

Quote:

Originally posted by Zifnab
You're right that it's convergent and it looks to converge to the value of 2

In the first place I'd suggest we all ought to say "sequence" instead of "series", for they are not really the same thing.

Having said this, at first glance I think you've hit the right idea (and a very nice one indeed), you've found a sequence with limit 2 whose terms are greater than our sequence:
a(n) > b(n) for all values of n (this is obvious by simple inspection of each pair of terms a_i, b_i)

I agree that this proves b(n) is convergent. However I don't think these sequences have both the same limit, in fact I believe the calculation I've outlined a few posts above points in the direction of a smaller limit for b_n in the vicinity of 1.7579...
The thing is, the way you've defined c_n, it is not always <2, take for instance:

b₄ = sqr(1 + c₄) -> c₄ = sqr(2 + sqr(3 + sqr(4))) = 2.058, aprox.

I wonder if now you've shown us the right direction to go we can come up with the value of the limit.
Jan 22nd, 2003, 03:02 AM
krtxmrtz

Oooops!

Quote:

Originally posted by myself
...you've found a sequence with limit 2 whose terms are greater than our sequence:
a(n) > b(n) for all values of n (this is obvious by simple inspection of each pair of terms a_i, b_i)

On re-reading this post I see I drew the above rash conclusion. I don't think it's that obvious that a_n > b_n

I still believe the limit is less than 2 but I don't see an easy way to prove it.
Jan 22nd, 2003, 07:40 AM
bugzpodder

MAK:

Set x_n to be the sequence.. it is quite easy to check that x_n is in fact increasing.. so if your limit is correct then x_n must be bounded by 2... so let us try to show x_n < 2.. this is equivalent to
n < (((((2^2-1)^2-2)^2-3)^2....)^2-(n-1))^2
this one I did by induction... easy to see
2 < (2^2-1)^2
assume it is true for n and let us prove it for n+1.. we have
n+1 < 2n < n^2 < n^2(n^2-2n+1)=n^2(n-1)^2=(n^2-n)^2
so if n < a then n+1 < (a^2-n)^2.. do you see it?? this will complete my proof...
Jan 24th, 2003, 07:02 PM
alkatran

No idea what ur talking about... but

Quote:

The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!

what does this mean? sounds interesting
Jan 24th, 2003, 08:14 PM
bugzpodder

if you pick a random rational number, it can be proven that after reducing it has a probability of 1/3 having an even denominator
Jan 25th, 2003, 04:40 PM
alkatran

if you pick a random rational number, it can be proven that after reducing it has a probability of 1/3 having an even denominator

I pick 2, so I .. reduce it? err... ok, 1. 1/1?? Grade 11 math terms plz :(
Jan 25th, 2003, 05:32 PM
bugzpodder

if you pick a random rational number, you have 1 in three chance of getting an even denominatored one. this is probability, ie ratio of success to total
Jan 25th, 2003, 06:11 PM
alkatran

Let me restate my question, I didn;t point out what I didin't understand.

if you pick a random rational number, you have 1 in three chance of getting an even denominatored one. this is probability, ie ratio of success to total

What do you mean.

1/3 3 is denominator correct? Would the random number NOT be an integer?
Jan 25th, 2003, 06:21 PM
bugzpodder

if you pick a random rational number, say 4/5 then its denominator is odd. a few other examples would be:
-8/6=-3/2 (even denominator)
3/7 (odd denominator)
13/2 (even denominator)
...

etc

now the chances of you getting an even denominator is 1/3
Jan 31st, 2003, 02:08 AM
krtxmrtz

Quote:

Originally posted by bugzpodder
MAK:

Set x_n to be the sequence.. it is quite easy to check that x_n is in fact increasing.. so if your limit is correct then x_n must be bounded by 2... so let us try to show x_n < 2.. this is equivalent to
n < (((((2^2-1)^2-2)^2-3)^2....)^2-(n-1))^2
this one I did by induction... easy to see
2 < (2^2-1)^2
assume it is true for n and let us prove it for n+1.. we have
n+1 < 2n < n^2 < n^2(n^2-2n+1)=n^2(n-1)^2=(n^2-n)^2
so if n < a then n+1 < (a^2-n)^2.. do you see it?? this will complete my proof...

That's a nice proof, but do you know how to effectively calculate the limit? (by means other than direct numerical calculation)
Jan 31st, 2003, 07:51 AM
bugzpodder

not a clue
Feb 2nd, 2003, 11:01 AM
NotLKH

Question #1

Quote:

if...
1 = sqr(1*k + sqr(2*k + sqr(3*k + ...sqr(n*k)...)))

What is k as n -> infinity?

Question #2

Quote:

if...
Targ = sqr(1*k + sqr(2*k + sqr(3*k + ...sqr(n*k)...)))

If Targ = k, then

What is k as n -> infinity, ?

Question #3

Quote:

if...
Targ = sqr(1*k + sqr(2*k + sqr(3*k + ...sqr(n*k)...)))

what does the plotting of k vrs targ look like, for each k when n approaches infinity? {My intuition suggests doing a log plot of k vrs Targ}

:)
Feb 3rd, 2003, 05:27 AM
krtxmrtz

We have a pending problem, how to calculate

lim (n -> infinity) sqr(1 + sqr(2 + sqr(3 + ...sqr(n)...)))

which looks dashed difficullt to solve. You're now proposing another 3 related questions in the same thread. Do you think you could establish some way to solve bugzpodder's original problem by solving yours first, e.g. as a special case?
Feb 3rd, 2003, 03:42 PM
NotLKH

Quote:

Originally posted by krtxmrtz
We have a pending problem, how to calculate

lim (n -> infinity) sqr(1 + sqr(2 + sqr(3 + ...sqr(n)...)))

which looks dashed difficullt to solve. You're now proposing another 3 related questions in the same thread. Do you think you could establish some way to solve bugzpodder's original problem by solving yours first, e.g. as a special case?

I figured if the problem was generalized even furthur, ie...

Quote:

Targ = sqr(1*k + sqr(2*k + sqr(3*k + ...sqr(n*k)...)))

Then studying the general properties and relationships between Targ and K when n is Some Large number {as in 2048}, then some insight might be gleaned.

I have no idea how to prove it, but I'm pretty certain that:

for some k,

Sqr(k) <= Targ <= Sqr(k) + 1

So, for Bugz's Original, where k = 1, then 1 <= Targ <= 2,

;)
Feb 3rd, 2003, 04:46 PM
bugzpodder

this is just like calculating the value of e. if you take its limit to more precision, you get more decimal points (i don't exactly know any algorithms for calculating e's digits but i am pretty sure there are a few). my point is, you cant just come up with a way to calculate this exact limit