Convergence of series

[Luca.Lussardi]

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 $\sum_(n=1)^(+\infty)\phi(a_n)$.

[Thomas]

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?

[Luca.Lussardi]

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.

[Thomas]

I have simply to check that it tends to 0, right???

[Luca.Lussardi]

That's right.

[Thomas]

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 $f(a_0)<a_0$ because $0<a_0<1$... this is sufficient for concluding that the succession tends to 0 thanks to the theory;

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!...

[Luca.Lussardi]

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.