Given a sequence $(x_n)$ and $x_{n+1}=x_n-nx_{n}^2$ for $n\geq1$. How to show the following? 
Given a sequence $(x_n)_n$ with $x_1 \in (0, 1)$ and $x_{n+1} = x_n - nx_n^2$ for $n \ge 1$, prove that the series $\sum_{n=1}^\infty x_n$ is convergent.

I am trying to solve this problem but could not proceed. Please give an idea to proceed.
 A: I would prove $\lim\limits_{n\to\infty}n^2 x_n=2$. Sketch: $x_n$ is positive (moreover, $0<nx_n<1/2$ for $n>1$ by induction); $y_n=1/x_n$ satisfies $y_{n+1}=y_n/(1-n/y_n)>y_n+n$, hence $n/y_n\to 0$ when $n\to\infty$, hence $(y_{n+1}-y_n)/n\to 1$, and thus $y_n/n^2\to 1/2$ by Stolz–Cesàro.
A: By looking at the first instance of the recursion, $x_2=x_1-x_1^2$, we see that $0<x_2\le \frac 14$. Then at the next step $x_3=x_2-2x_2^2$, we see that $0<x_3\le \frac 18$ for $0<x_2\le \frac 14$. It is beginning to look good, the terms seem to tend to zero as fast as powers of two, but if we go one step further, we only find $0<x_4\le \frac18-\frac3{64}=\frac5{64}(>\frac1{16})$. So comparison with $\sum \frac1{2^n}$ (or any other geometric series) will not work, the $x_n$ seem to decrease nicely but seemingly too slow for a geometric series. 
So perhaps compare the bounds $1,\frac14,\frac18,\frac5{64},\ldots$ with the slower $\frac c{n^2}$ instead? 
For a proof by induction, assume $0<x_n\le \frac c{n^2}$ and $n\ge 2c$. Then in particular $x_n\le \frac 1{2n}$ and as $x\mapsto x-nx^2$ is strictly increasing on $[0,\frac1{2n}]$, we can conclude that $$\tag10<x_{n+1}\le \frac c{n^2}-n\cdot\frac{c^2}{n^4}=c\cdot \frac{n-c}{n^3}.$$
Note that with $c=\frac 54$ and for $n\ge3$, 
$$ (n-c)(n+1)^2-n^3=(2-c)n^2+(1-2c)n-c.$$
In order have $\frac{n-c}{n^3}\le \frac1{(n+1)^2}$ as we want that to continue $(1)$ as desired, we want the right side to be non-negative. So we better pick $c=2$. Now what we found amounts to

Suppose $n\ge 4$ and $0<x_n\le\frac 2{n^2}$. Then $0<x_{n+1}\le\frac 2{(n+1)^2}$.

Fortunately, we already verified $0<x_4\le\frac5{64}<\frac2{4^2}$ and so conclude that $0<x_n\le\frac 2{n^2}$ and the convergence of $\sum x_n$ follows from that of $\sum \frac1{n^2}$.
