We have $x_{n+1}=x_n+\frac{1}{n^2x_n}$. I am trying to see how the sequence convergence depends on the first term $x_1 > 0$.
I've been calculating sequence terms for different initial ones and working out the recurrence relation, but I haven't made much progress.
Can someone give me a hint? Is there anything known about sequences of this type?
The sequence converges for every $x_1>0$. To see why, first rewrite $$ x_{n+1}-x_n = \frac{1}{n^2 x_n} > 0 \tag{1} $$ (it is immediate to show by induction that $x_n > 0$ for all $n$). This shows that the sequence is monotone increasing, and therefore by monotone convergence either converges or diverges to $\infty$.
Now, summing (1) from $n=1$ to $N$, we get $$ 0 < x_{N+1}-x_1 = \sum_{n=1}^N (x_{n+1}-x_n) = \sum_{n=1}^N\frac{1}{n^2 x_n}\leq \frac{1}{x_1}\sum_{n=1}^N\frac{1}{n^2}< \frac{1}{x_1}\sum_{n=1}^\infty\frac{1}{n^2}$$ and therefore the series $(x_n)_n$ is bounded. By the above, it it therefore convergent to some $L>0$ (the value of $L$ may and will depend on $x_1$).