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.
Questions:
- 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.
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!