# Bounding an implicitly defined sequence

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

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

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