I have a sequence $\lambda_0,\lambda_1,\ldots,$ which is defined recursively as

$$ \lambda_0 = \frac{1}{2},$$

and

$$\lambda_{k+1} = \max_{\lambda\in [1,b]} \left(\frac{1}{2\lambda}\prod_{0\leq j\leq k}\left(\frac{\lambda-\lambda_j}{\lambda+\lambda_j}\right)^2\right), \qquad k\geq 0,$$

where $b>1$. I would like to find a relatively sharp upper bound for $\lambda_k$. Numerically, it seems that $\lambda_k$ is bounded by something like,

$$\lambda_k \leq \mathcal{O}\left(e^{-8k/\log(b)}\right). $$

For instance, the following very simple argument gives an upper bound that is very weak and I'm seeking techniques to do better:

Note that we have, for $k\geq 0$,

$$ \lambda_{k+1} \leq \max_{\lambda\in[1,b]} \left(\frac{1}{2\lambda}\prod_{0\leq j\leq k-1}\left(\frac{\lambda-\lambda_j}{\lambda+\lambda_j}\right)^2\right)\max_{\lambda\in[1,b]}\left(\frac{\lambda - \lambda_k}{\lambda+\lambda_k}\right)^2\leq \lambda_k \left(\frac{b-1}{b+1}\right)^2$$

and hence,

$$\lambda_k \leq \left(\frac{b-1}{b+1}\right)^{2k}\lambda_0 =\frac{1}{2}\left(\frac{b-1}{b+1}\right)^{2k},\qquad k\geq0.$$

I've already asked this on the Math Stack exchange, but it didn't receive any response:

http://math.stackexchange.com/questions/372661/bounding-an-implicitly-defined-sequence

Thank you in advance.