首先 $a_{n+1} > a_n$, 故 $\{a_n\}$ 是严格单调递增的数列.
将递推公式两边取倒数,
\[
\frac{1}{a_{n+1}}=\dfrac{1}{a_n+\frac{a_n^2}{n^2}}=\frac{1}{a_n}\cdot\frac{1}{1+\frac{a_n}{n^2}}=\frac{1}{a_n}\cdot\Bigl(1-\dfrac{\frac{a_n}{n^2}}{1+\frac{a_n}{n^2}}\Bigr)=\frac{1}{a_n}-\frac{1}{n^2+a_n},
\]
即
\[
\frac{1}{a_n}-\frac{1}{a_{n+1}}=\frac{1}{n^2+a_n}.
\]
于是
\[
\frac{1}{a_1}-\frac{1}{a_{n+1}}=\sum_{k=1}^{n}\Bigl(\frac{1}{a_k}-\frac{1}{a_{k+1}}\Bigr)=\sum_{k=1}^{n}\frac{1}{k^2+a_k}.
\]
这推出
\[
\begin{split}
\frac{1}{a_{n+1}}&=\frac{1}{a_1}-\sum_{k=1}^{n}\frac{1}{k^2+a_k}\\
&=\frac{1}{a_1}-\biggl[\frac{1}{1^2+a_1}+\frac{1}{2^2+a_2}+\frac{1}{3^2+a_3}+\sum_{k=4}^{n}\frac{1}{k^2+a_k}\biggr]\\
& > \frac{1}{\frac{2}{5}}-\biggl[\frac{1}{1+\frac{2}{5}}+\frac{1}{4+\frac{12}{25}}+\frac{1}{9+\frac{303}{625}}+\sum_{k=4}^{n}\frac{1}{k^2+a_4}\biggr]\\
&=\frac{5}{2}-\frac{12365}{11856}-\sum_{k=4}^{n}\frac{1}{k^2+a_4}\\
& > \frac{5}{2}-\frac{12365}{11856}-\sum_{k=4}^{\infty}\frac{1}{k^2}\\
&=\frac{17275}{11856}-\frac{\pi^2}{6}+\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}\\
&=\frac{100237}{35568}-\frac{\pi^2}{6}
\end{split}
\]
取 $\pi$ 为
>> pi()
out> 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679
------------------------
>> 100237/35568-pi^2/6
in> 100237/35568-3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679^2/6
out> 1.1732451954099831901582640956121844374117293610514111888951210328882021562274783884245807260109205571
------------------------
>> 1/1.1732451954099831901582640956121844374117293610514111888951210328882021562274783884245807260109205571
in> 1/1.1732451954099831901582640956121844374117293610514111888951210328882021562274783884245807260109205571
out> 0.8523367527199259137223649929392592715446226054869969414563359727960636278088041445804898332626884568
------------------------
因此
\[
a_{n+1} < 0.852336752719925913722364993,\quad\forall\ n\in\mathbb{Z}^+
\]