Calculus problem involving recurrent sequences I found this interesting problem in a Romanian magazine. I managed to show that the sequence is convergent, but I couldn't find the limit. Let $(a_n)_{n \ge 1}$ a sequence defined as follows: 
$$a_1 \gt 0$$$$ a_{n+1}= \frac {na_n} {n+a_n^2},\forall n \ge 1.$$
calculate $$ \lim _ {n \to \infty} a_n.$$
Later edit. I showed the convergence in two steps: 1) I calculated $a_{n+1}/a_n$ which is less than 1, so the sequence is strictly dereasing; 2)The terms are between $a_1$ and $0$, so it is bounded. By Weierstrass' Theorem, the sequence is convergent.
 A: A possible approach:


*

*Prove convergence. 


*

*By induction, show that $a_n > 0$ for all $n\geq 1$.

*Observe then that $0 < \frac{a_{n+1}}{a_n} = \frac{n}{n+a)n^2} < 1$, so that the sequence is decreasing.

*Invoke the monotone convergence theorem: the sequence converges to some $\ell\in[0,a_1)$.


*Find the limit. Suppose by contradiction that $\ell > 0$ (the proof can be easily adapted to be done without contradiction, but then one has to be careful when taking equivalents to avoid the "equivalent to $0$" issue. So I am going for a proof by contradiction for simplicity). Observe that
$$
\ln \frac{a_{n+1}}{a_1} = \ln \prod_{k=1}^n \frac{a_{k+1}}{a_k}
= \ln \prod_{k=1}^n \frac{1}{1+\frac{a_k^2}{k}}
= -\sum_{k=1}^n\ln\left(1+\frac{a_k^2}{k}\right).
$$
By theorems of comparison of series, and since the sequence converges to $\ell$ (so that $\frac{a_k^2}{k}\xrightarrow[k\to\infty]{}0$) we have that 
$$
\sum_{k=1}^n\ln\left(1+\frac{a_k^2}{k}\right) \operatorname*{\sim}_{n\to\infty}
\sum_{k=1}^n \frac{a_k^2}{k}\operatorname*{\sim}_{n\to\infty}
\ell^2 \sum_{k=1}^n \frac{1}{k}\operatorname*{\sim}_{n\to\infty}
\ell^2 \sum_{k=1}^n \frac{1}{k}
\operatorname*{\sim}_{n\to\infty}
\ell^2 \ln n
$$
(where for the equivalent we used that $\ell > 0$, as otherwise "equivalent to $0$" would not be formally correct in that case).
But then,
$$
\ln \frac{a_{n+1}}{a_1} \operatorname*{\sim}_{n\to\infty} -\ell^2 \ln n \xrightarrow[n\to\infty]{} -\infty
$$
i.e. 
$$
\ln a_{n+1} \xrightarrow[n\to\infty]{} -\infty
$$
and by unicity of the limit this means that $\ell = 0$. Contradiction.
So $$\boxed{\ell=0.}$$

For the sake of illustration, here is a plot of the first 100 elements of the sequence, obtained with Mathematica (for $a_1\in\{1,2,3\}$):

A: (This is a massive revision
of my original answer,
which sits lonely at the end.)
It converges to zero.
I am quite sure that
the true order of $a_n$
is
$\dfrac1{\sqrt{2} \sqrt{c + \log(n)}}
$.
Since
$\frac{a_n}{a_{n+1}}
=1+a_n^2/n
\gt 1
$,
$a_n>a_{n+1}$
so that
$\lim a_n$
exists and is non-negative.
To show that
$\lim a_n=0$,
$\begin{array}\\
\dfrac{a_n}{a_{n+k}}
&=\prod_{j=1}^k \dfrac{a_{n+j-1}}{a_{n+j}}\\
&=\prod_{j=1}^k (1+a_{n+j}^2/(n+j))\\
&>\sum_{j=1}^k \dfrac{a_{n+j}^2}{n+j}\\
\end{array}
$
If $\lim a_n >0$,
then
$\sum_{j=1}^k \dfrac{a_{n+j}^2}{n+j} \to \infty$
as $k \to \infty$
which makes
$\dfrac{a_n}{a_{n+k}} \to \infty$,
so
$\dfrac{a_{n+k}}{a_n} \to 0$,
a contradiction.
To estimate how
$a_n \to 0$,
suppose
$a_n \approx 1/f(n)$,
then
$\begin{array}\\
\dfrac{a_n}{a_{n+1}}
&\approx\dfrac{f(n+1)}{f(n)}\\
&\approx\dfrac{f(n)+f'(n)}{f(n)}\\
&=1+\dfrac{f'(n)}{f(n)}\\
\end{array}
$
and
$1+\dfrac{a_n^2}{n}
\approx 1+\dfrac{1}{nf(n)^2}
$
so that we want
$\dfrac{1}{nf(n)^2}
=\dfrac{f'(n)}{f(n)}
$
or
$f'(n) f(n) = \dfrac1{n}
$.
According to Wolfy,
the solution to this is
$f(x)
= \sqrt{2} \sqrt{c + \log(x)}
$,
so that
$a_n
\approx \dfrac1{\sqrt{2} \sqrt{c + \log(n)}}
$.

Here is my original solution:

It goes to zero because
$a_{n+1}/a_n =1/(1+a_n^2/n)
<1-a_n^2/(2n)$.
If $\lim a_n >0$,
then
$\sum a_n^2/(2n)$
diverges,
which makes
$\prod a_{n+1}/a_n \to 0$,
a contradiction.
