Mathematical Reflections 2015, Issue 1 - Problem U327

Problem:
Let $(a_n)_{n \geq 0}$ be a sequence of real numbers with $a_0=1$ and $$a_{n+1}=\dfrac{a_n}{n^2a_n+a^2_n+1}.$$
Find the limit $\displaystyle \lim_{n \to \infty} n^3 a_n$.

Proposed by Khakimboy Egamberganov.

Solution:
Observe that $a_{n+1}<\dfrac{a_n}{n^2a_n}=\dfrac{1}{n^2}$, so $\displaystyle \sum_{n=0}^{\infty} a_n$ converges by the Comparison Test. Let $$\sum_{n=0}^{\infty} a_n=c \in \mathbb{R}.$$
We have $$\dfrac{1}{a_{n+1}}=n^2+a_n+\dfrac{1}{a_n}.$$ So, $$\begin{array}{rcl} \dfrac{1}{a_1}&=&0^2+a_0+\dfrac{1}{a_0} \\ \dfrac{1}{a_2}&=&1^2+a_1+\dfrac{1}{a_1} \\ \vdots & \vdots & \vdots \\ \dfrac{1}{a_n}&=&(n-1)^2+a_{n-1}+\dfrac{1}{a_{n-1}},\end{array}$$ and summing the two columns, we get $$\dfrac{1}{a_n}=\dfrac{(n-1)n(2n-1)}{6}+\sum_{k=0}^{n-1} a_k+1.$$
Hence, $$\dfrac{1}{n^3a_n}=\dfrac{(n-1)n(2n-1)}{6n^3}+\dfrac{\sum_{k=0}^{n-1} a_k+1}{n^3}.$$ Since $$\displaystyle \lim_{n \to \infty} \dfrac{\sum_{k=0}^{n-1} a_k+1}{n^3}=\lim_{n \to \infty} \dfrac{c+1}{n^3}=0,$$ we get $$\lim_{n \to \infty} \dfrac{1}{n^3 a_n}=\lim_{n \to \infty} \dfrac{(n-1)n(2n-1)}{6n^3}=\dfrac{1}{3},$$ so $$\lim_{n \to \infty} n^3 a_n=3.$$