I have problems finding limits of recursive sequence like: $a_1 = 5$ $a_{n+1} = \frac{(a_n)^2}{10n + 3} + 1$
I know that intuitively the limit will be $1$, since $\frac{(a_n)^2}{10n + 3}$ tends to 0, but I have problems proving that.
Hints would be appreciated.
On
First to all we'll need to show that $a_n$ is monotone, in particular decreasing by induction.
It's simple verify that $a_2<a_1$ and $a_n\geq 1$ for every $n$ because the term $\frac{a^2_n}{10n+3}$ is always $\geq 0$, now suppose $a_n<a_{n-1}$ then $a_n^2<a_{n-1}^2$ and $$ a_{n+1}=\frac{a_n^2}{10(n+1)+3}+1<\frac{a_{n-1}^2}{10(n+1)+3}+1<\frac{a_{n-1}^2}{10n+3}+1=a_n $$
Now $a_n$ is decreasing and lower bounded by $1$ so $a_n$ has got a limit: $\lim_{n\rightarrow +\infty}a_n=l\in\mathbb R$. Then $$ l=\lim_{n\rightarrow +\infty}a_{n+1}=\lim_{n\rightarrow +\infty}\frac{a_n^2}{10n+3}+1=\frac l{+\infty}+1=0+1=1 $$ then $l=1$.
It should be clear that $a_n >0$ for all n.
Now prove, by induction, that $(a_n)$ is decreasing.
Hence $(a_n)$ is monotonic and bounded and therefore convergent. If $a$ is the limit of $(a_n)$ then $ \frac{(a_n)^2}{10n + 3} + 1 \to 1$ as $n \to \infty$, hence $a_{n+1} \to 1$ as $n \to \infty$.