$\displaystyle \sum_{k=1}^{infinity}a_{k}$ where $\displaystyle a_{k+1}=cos(a_{k}) $for k>=1

is it converges absolutly ,or with condition(dont know the english term),

or is it diverges?

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- Mar 2nd 2011, 01:02 AMtransgalacticconvergence of recursive series mmn11 2d
$\displaystyle \sum_{k=1}^{infinity}a_{k}$ where $\displaystyle a_{k+1}=cos(a_{k}) $for k>=1

is it converges absolutly ,or with condition(dont know the english term),

or is it diverges? - Mar 2nd 2011, 02:24 AMTed
Are you given the value of $\displaystyle \displaystyle a_0$ ?

- Mar 2nd 2011, 03:26 AMtransgalactic
no

- Mar 2nd 2011, 05:28 AMchisigma
The recursive relation can be written as...

$\displaystyle \displaystyle \Delta_{k}= a_{k+1} - a_{k} = \cos a_{k} - a_{k} = f(a_{k})$ (1)

The function f(x) is represented here...

http://digilander.libero.it/luposabatini/MHF110.bmp

There is only one 'attractive fixed point' at $\displaystyle x_{0}= .739085133215...$ and because $\displaystyle \forall x$ is $\displaystyle |f(x)|<|2\ (x_{0}-x)|$ [red line...] any possible sequence converges at $\displaystyle x_{0}$. Because is $\displaystyle |f(x)|>|x_{0}-x|$ [blue line...] however the convergence will be not 'monotonic' but 'oscillating'. Of course is $\displaystyle \displaystyle \lim_{k \rightarrow \infty} a_{k} \ne 0$ the series $\displaystyle \displaystyle \sum_{k=1}^{\infty} a_{k}$ diverges...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Mar 3rd 2011, 06:16 AMtransgalactic
its a recursive series by what test should i test the absolute value?

- Mar 3rd 2011, 07:20 AMchisigma
The general term donsn't tends to 0 if k tends to infinity, so that the series diverges... it is interesting to valuate the series $\displaystyle \sum_{k=1}^{\infty} a_{k}$ where the $\displaystyle a_{k}$ satisfy the recursive relation...

$\displaystyle \displaystyle a_{k+1}= \sin a_{k}$ (1)

The (1) can be written as...

$\displaystyle \displaystyle \Delta_{k}= a_{k+1} - a_{k} = \sin a_{k}- a_{k}= f(a_{k})$ (2)

The function f(x) is represented here...

http://digilander.libero.it/luposabatini/MHF111.bmp

There is only one 'attractive fixed point' at $\displaystyle x_{0}=0$ and it is clear that is...

$\displaystyle \displaystyle \lim_{k \rightarrow \infty} a_{k}=0$ (3)

... no matter which is $\displaystyle a_{0}$ and the convergence is 'monotonic', so that the series has only positive or negative terms. The (3) tells us that the series may converge, but the question is: does it converge or not?... the question isn't trivial... for example the ratio test fails because is...

$\displaystyle \displaystyle \lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_{k}} = \lim_{k \rightarrow \infty} \frac{\sin a_{k}}{a_{k}} = 1$ (4)

Does someone have some 'good idea'?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Mar 3rd 2011, 09:36 AMtransgalactic
i have to foolow the schematics

first make test on absolute value then go to the original

here it doesnt work

?? - Mar 4th 2011, 03:19 AMtransgalactic
ohhh sorry didnt notices that a1=1

does it help? - Mar 4th 2011, 03:36 AMHallsofIvy
It is still true that $\displaystyle a_k$ does NOT go to 0 so the series does NOT converge, conditionally or absolutely.

- Mar 5th 2011, 12:40 AMtransgalactic
but we dont have expression for a_n

its recursive - Mar 5th 2011, 01:29 AMHallsofIvy
So what is your point? We don't need an expression for $\displaystyle a_n$. cosine of any number is between -1 and 1 and cosine of any number between -1 and 1 is larger than .5. The terms do NOT go to 0 so the series cannot converge.

- Mar 5th 2011, 04:21 AMtransgalactic
but its a recursive expresion we dont have an expression of a_n with "n" only variable

we cant say to what a_n goes

because it is recursive - Mar 5th 2011, 04:26 AMHallsofIvy
You can keep saying that as much as you like, but it is not relevant to the question. You don't

**need**to know an explicite formula for $\displaystyle a_n$ to see that this series does not converge. - Mar 5th 2011, 09:33 PMtransgalactic
but i need to prove it mathmatickly

what kind of limit should i write? - Mar 6th 2011, 03:50 AMHallsofIvy
Perhaps I am misunderstanding

**what**you are trying to prove. I thought you were trying to determine whether or not the series $\displaystyle \sum a_n$ converged and prove it.

One of the very first theorems about infinite series is that if the series $\displaystyle \sum a_n$ converges, then the sequence $\displaystyle \{a_n\}$ must go to 0. The contrapositive of that is that if the sequence does not go to 0, then the series does not converge.

Now, what exactly are you trying to prove?