Let $(x_n)_{n\geq 1} \subset \mathbb{R}$ such that $x_1>0$ and $$x_{n+1}=\frac{1}{2}(x_n+\sqrt{x_n^2+a_n}), \quad \forall n \geq 1$$ Determine if $x_n$ converges or diverges when

a) $a_n=\frac{1}{\sqrt{n}}$

b) $a_n=\frac{b_n^2}{(1+b_1^2+b_2^2+\dots+b_n^2)^2}$, where $(b_n)_{n \geq 1} \subset \mathbb{R}$

It is quite easy to see that $x_n>0, \forall n\geq 1$ and $(x_n)$ is strictly increasing. I didn't succeed in finding a bound for a) and I think that $(x_n)$ diverges, but couldn't prove it.


Note that

$$x_{n+1} - x_n = \frac{1}{2}\bigl(\sqrt{x_n^2 + a_n} - x_n\bigr) = \frac{a_n}{2(\sqrt{x_n^2+a_n} + x_n)}.$$

Hence, if $(a_n)$ is a bounded positive sequence, $(x_n)$ converges if and only if

$$\sum_{n = 1}^{\infty} a_n < +\infty.$$


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