$a_{n+1}=a_n-(a_n)^2$ determine the value of $a_0$ thak makes $\lim_{n\to\infty}a_n=0$
the book says that the answer is $a_0\in[0,1]$
I tried to find $a_n$ in terms of n but failed. Is this approach correct or are there any other ways to solve this problem?
Note that $a_n-a_{n+1}=a_n^2\ge 0$, so the sequence is decreasing. This proves that $a_n\ge 0$ for all $n\in\Bbb N$.
On the other hand, if for some $n$ we have $a_n>1$ then $a_{n+1}<0$ so $a_n\le 1$ for all $n\in\Bbb N$.
If $a_n=0$ for some $n$ then all the following terms are $0$. The same if $a_1=1$.
Otherwise, dividing by $a_n$ we get $$0<\frac{a_{n+1}}{a_n}=1-a_n<1-a_1<1$$ so $a_n\to 0$.