# Sequence convergence and limit value depending on initial term

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?

• Presumably, $x_1 > 0$? – Clement C. Apr 9 '18 at 18:51

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 $x_1$).
• Similarly for $x_1 < 0$, consider $y_n = -x_n$. Then $y_n$ satisfies the same recurrence, and $y_1 > 0$, giving that $y_n$ converges, hence $x_n$ does also. – B. Mehta Apr 9 '18 at 19:17
• @Asix: In that case, I'd argue it makes much more sense (and could be easier) to look for a dependence on $x_2$. Note that $x_2 = x_1+1/x_1 \geq 2$ for any $x_1>0$, so in particular any initial choice for $x_1$ of either $a$ and $1/a$ would lead to the same limit. – Clement C. Apr 9 '18 at 20:00