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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?

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

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    $\begingroup$ 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. $\endgroup$ – B. Mehta Apr 9 '18 at 19:17
  • $\begingroup$ But there is no way to find a direct relation between the first term and the limita, no? $\endgroup$ – Asix Apr 9 '18 at 19:18
  • $\begingroup$ There may be. Is that what you mean? Your question asks "how the sequence convergence depends on the first term" -- which I interpreted as "whether the series converges or not, depending on the first term" (that's what "convergence" means). Did you mean instead "How the value of the limit depends on the first term"? $\endgroup$ – Clement C. Apr 9 '18 at 19:20
  • $\begingroup$ I was thinking about both...sorry for the misunderstanding. $\endgroup$ – Asix Apr 9 '18 at 19:25
  • $\begingroup$ @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. $\endgroup$ – Clement C. Apr 9 '18 at 20:00

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