Given the sequence $a_{n+1}=a_n(1-\sqrt{a_n})$ with $a_1\in(0,1)$ show that $b_n=a_1^2+a_2^2+\cdots+a_n^2$ is upper bounded by $a_1$.

Question

Consider the sequence $(a_n)_{n \in \mathbb{N}^*}$ with:

$$a_1 \in (0, 1)$$

$$a_{n+1} = a_n(1 – \sqrt{a_n})$$

and also the sequence $(b_n)_{n \in \mathbb{N}^*}$ with:

$$b_n = a_1^2 + a_2^2 + \cdots + a_n^2$$

I have to show that the sequence $(b_n)$ is upper bounded by $a_1$.

So far I managed to show that $a_n \in (0, 1), \forall a_1 \in (0, 1)$ by induction and also that $(a_n)$ is strictly decreasing. I showed these two things in the hopes that it would somehow help me show what is asked. But I didn't find a way of doing that. It is also quite obvious that the sequence $(b_n)$ is strictly increasing, but yet again I don't know if that will help me in showing that it is upper bounded by $a_1$. So how should I approach this?

Answer 1

$a_{n+1}=a_n-a_n^{3/2} \leq a_n-a_n^{2}$ because $a_n^{2} <a_n^{3/2}$. Hence $a_n^{2} \leq a_n-a_{n+1}$. Sum this from $n=1$ to $m$ to get $b_m \leq a_1-a_{m+1} \leq a_1$.