Consider the recursive relation $a_{n+1}:=a_n+\cfrac{a_n^2}{n^2}$. The existence of $\lim_n a_n$ depends on the initial value $a_1$. For instance:

If $a_1=1$, then $a_n=n$ and the sequence is divergent.

If $a_1=0$, then $a_n=0$ and the sequence is convergent.


  • Numerical calculation shows that if $a_1\in(-2,1)$, then it is convergent. Is that right? How to prove that, and can we find the limit?
  • How about $a_1\in \mathbb{C}$ ?

P.S: I found this related to Göbel's Sequence.

  • $\begingroup$ it should converge at least for every $z \in \Bbb C$ with $|z|<1$ $\endgroup$ – mercio Oct 14 '17 at 7:30
  • $\begingroup$ numerically it also seems to converge when $z=i$. $\endgroup$ – Ewan Delanoy Oct 14 '17 at 7:43
  • $\begingroup$ also if $a_1 = -1-x$ or $a_1=x$ then $a_2 = x+x^2$ in both cases so there is a huge symmetry. $\endgroup$ – mercio Oct 14 '17 at 7:56
  • $\begingroup$ also it looks like any $z$ with $|z| \ge 4$ would give a sequence that goes to infinity. $\endgroup$ – mercio Oct 14 '17 at 8:24
  • $\begingroup$ @mercio I found it looks the same as Julia set $Re=0.25,Im=0$. $\endgroup$ – GalAster Oct 14 '17 at 8:27

For $-1<a_1<0$ we can see that $a_n<0$ and $a$ increases.

The case $-2<a_1<-1$ should be similar to the following case, but I have no a proof.

Let $0<a_1<1$.

By induction easy to show that $a_n<na_1$.

Thus, there is $k$, for which $a_{k}<k-1$.

Also, we have: $$\frac{1}{a_n}-\frac{1}{a_{n+1}}=\frac{a_n}{n^2a_{n+1}}<\frac{1}{n^2}<\frac{1}{n(n-1)}.$$ Thus, for all $n>k$ we obtain: $$\frac{1}{a_k}-\frac{1}{a_n}=\sum_{i=k}^{n-1}\left(\frac{1}{a_{i}}-\frac{1}{a_{i+1}}\right)<$$ $$<\sum_{i=k}^{n-1}\left(\frac{1}{i-1}-\frac{1}{i}\right)=\frac{1}{k-1}-\frac{1}{n-1}<\frac{1}{k-1}.$$ Id est, $$\frac{1}{a_n}>\frac{1}{a_k}-\frac{1}{k-1}$$ and since $$\frac{1}{a_k}-\frac{1}{k-1}>0,$$ we are done!

  • 2
    $\begingroup$ The sequence for $a_1=x$ is the same (except for the first term) as for $a_1=-x-1$. Therefore, $-2<a_1<-1$ behaves just as $0<a_1<1$ does. $\endgroup$ – Hagen von Eitzen Oct 14 '17 at 10:40

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