Prove the existence of limit of $x_{n+1}=x_n+\frac{x_n^2}{n^2}$ The problem is: Let $\{x_n\}$ be a sequence such that $0&lt;x_1&lt;1$ and $x_{n+1}=x_n+\dfrac{x_n^2}{n^2}$. Prove that there exists the limit of $\{x_n\}$.

It is easy to show that $x_n$ is increasing, but I cannot prove it is bounded to show the existing of limit. Anyone have any ideas?
Let $x_1 = t \in (0,1)$. For any $n \ge 1$, it is easy to see $x_n \ge 0$.
Since $e^y \ge 1 + y$ for all $y$, we have
$$x_n = x_1 \prod_{k=1}^{n-1} \frac{x_{k+1}}{x_k} = t \prod_{k=1}^{n-1} \left(1 + \frac{x_{k}}{k^2}\right) \le t\exp\left(\sum_{k=1}^{n-1} \frac{x_k}{k^2}\right)\tag{*1}$$

Notice


$x_1 \le t$.
if $x_n \le tn$, then $x_{n+1} \le n t + t^2 \le (n+1)t$.


By induction, we have $x_n \le n t$ for all $n$. 

Substitute this into $(*1)$, we get

$$x_n \le t \exp\left(t\sum_{k=1}^{n-1}\frac{1}{k}\right)
\le t\exp\left(t(\log n + \gamma)\right)
\le  e^\gamma n^t$$
where $\gamma$ is 
Euler-Mascheroni constant.

Substitute above into $(*1)$ again, we find
$$x_n \le t \exp\left(e^{\gamma} \sum_{k=1}^{n-1}\frac{1}{k^{2-t}}\right)
 \le t \exp\left( e^\gamma \zeta(2-t)\right) &lt; \infty$$
where $\zeta(t)$ is the Riemann Zeta function.
The sequense is obviously monotone increasing since $x_{n+1}&gt;x_n$. To prove that is converging we should show that is bounded.
Expanding $x^n$ and doing the addition we take:
$$
\eqalign{
x_{n+1} &amp;= x_n + {x_n^2 \over n^2} \cr
x_{n} &amp;= x_{n-1} + {x_{n-1}^2 \over {(n-1)}^2} \cr
x_{n-1} &amp;= x_{n-2} + {x_{n-2}^2 \over {(n-2)}^2} \cr
&amp;\cdots \cr
x_2 &amp;= x_1 + {x_1^2 \over 1^2} \cr
&amp;\rightarrow \cr
x_{n+1} &amp;= x_1+ {x_{1}^2 \over {1}^2}+ {x_{2}^2 \over {2}^2}+\cdots+ {x_{n-1}^2 \over {(n-1)}^2}+ {x_{n}^2 \over {n}^2} \cr
&amp;&lt; 1+ {2 \over {1}^2}+ {2 \over {2}^2}+ \cdots+ {2 \over {(n-1)}^2}+ {2 \over {n}^2} \cr
&amp;&lt;1+ {2 \pi^2 \over 6}
}
$$