$\displaystyle \sum_{k=1}^{infinity}a_{k}$ 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?
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...
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
$\displaystyle \chi$ $\displaystyle \sigma$
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...
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
$\displaystyle \chi$ $\displaystyle \sigma$
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?