A limit problem about $a_{n+1}=a_n+\frac{n}{a_n}$

Solution 1:

We have: $$ a_n(a_{n+1}-a_n) = n $$ so: $$ a_{n+1}^2-a_{n}^2 = a_{n+1}(a_{n+1}-a_n) + n = n\left(1+\frac{a_{n+1}}{a_n}\right)=2n+\frac{n^2}{a_n^2}$$ and: $$ a_{N+1}^2-a_1^2 = N(N+1)+\sum_{n=1}^{N}\frac{n^2}{a_n^2} $$ from which $a_{N+1}\geq \sqrt{N(N+1)}$ and $a_n\geq \sqrt{(n-1)n}$.

If we plug this inequality back into the previous line, we get: $$\begin{eqnarray*} a_{N+1}^2 &\leq& N(N+1)+a_1^2+\frac{1}{a_1^2}+\sum_{n=2}^{N}\frac{n}{n-1}\\&=& (N+1)^2+\left(a_1-\frac{1}{a_1}\right)^2+H_{N-1}.\end{eqnarray*} $$ The process continues by keep turning lower/upper bounds into tighter upper/lower bounds.

Can you check it proves your statement?