Convergence of series
Let $\alpha \in (0,1)$ and consider the real sequence $a_0=\alpha$, $a_(n+1)=(2a_n^2)/(1+a_n)$, $n \ge 1$. Let $\phi(x)=1/(log(x+1/x))$, for $x>0$. Find the behaviour of $\sum_(n=1)^(+\infty)\phi(a_n)$.
Risposte
Ok, even if an application of De l'Hopital rule can easily show that the limit is strictly less than $1$ ($0$ if I remember right...) so that the serie converges by ratio criterion.
Ok... $a_n$ tends to 0 because:
- it's a succession of the type $a_(n+1)=f(a_n)$, with $f(x)=(2x^2)/(1+x)$, wich is a monotone increasing function... the solutions of the equation $f(x)=x$ are 0 and 1, furthermore $f(a_0)
done that, I'd like somebody to compute that limit for me!
... (it's simply the limit of $a_(n+1)/a_(n)$)... if it tends to $1^(+)$ the sum diverges, if it tends to $a<1$, it converges, if it tends to 1^(-), well... I hope it doesn't!...
- it's a succession of the type $a_(n+1)=f(a_n)$, with $f(x)=(2x^2)/(1+x)$, wich is a monotone increasing function... the solutions of the equation $f(x)=x$ are 0 and 1, furthermore $f(a_0)
done that, I'd like somebody to compute that limit for me!
... (it's simply the limit of $a_(n+1)/a_(n)$)... if it tends to $1^(+)$ the sum diverges, if it tends to $a<1$, it converges, if it tends to 1^(-), well... I hope it doesn't!...
That's right.
I have simply to check that it tends to 0, right???
No, you are not far from the solution, hence the limit you have posted solves the question, if one find the exact behaviour of $a_n$ of course... but this is not much complicated.... it's enuogh a point fix argument.
could it be useful to compute:
$lim_(x->0)log(x+1/x)/log((2x^2)/(1+x)+(1+x)/(2x^2))$
???
otherwise, I could try to find the asymptotic behaviour of $a_n$...
am I really far from the solution?
$lim_(x->0)log(x+1/x)/log((2x^2)/(1+x)+(1+x)/(2x^2))$
???
otherwise, I could try to find the asymptotic behaviour of $a_n$...
am I really far from the solution?
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