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

Let $a_{n+1}=a_n+\frac{n}{a_n}$ and $a_1>0$. Prove $\lim\limits_{n\to \infty} n(a_n-n)$ exists.

In my view, maybe we can use $${a_{n + 1}} = {a_n} + \frac{n}{{{a_n}}} \Rightarrow {a_{n + 1}} - \left( {n + 1} \right) = \left( {{a_n} - n} \right)\left( {1 - \frac{1}{{{a_n}}}} \right).$$ And then $${a_n} - n = \left( {{a_1} - 1} \right)\prod\limits_{k = 1}^{n - 1} {\left( {1 - \frac{1}{{{a_k}}}} \right)} .$$ By Stolz formula, we have \begin{align*} &\mathop {\lim }\limits_{n \to \infty } n\left( {{a_n} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\frac{1}{{{a_n} - n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\frac{1}{{{a_{n + 1}} - \left( {n + 1} \right)}} - \frac{1}{{{a_n} - n}}}}\\ = &\mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{{a_n} - {a_{n + 1}} + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{ - \frac{n}{{{a_n}}} + 1}}\\ = &\mathop {\lim }\limits_{n \to \infty } {a_n}\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]. \end{align*} And how can we continue?

• Think about what you know about $a_n$ (substitute) – Julian Rachman Oct 23 '15 at 7:35

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?

1) You can clearly see from what you have developed that $a_n\geq n$ (equality for $a_1=1$. For the rest we assume $a_1=1+\alpha>1$).
2) It is also easy to see that $a_{n}-n$ is decreasing. So if $a_1=\alpha+1$ then $a_n-n<\alpha$.
3) Consider the product $\prod\limits_{k = 1}^{n - 1} {\left( {1 - \frac{1}{{{a_k}}}} \right)}$, we can show it converges to zero using following inequalities: $$\sum\limits_{k = 1}^{n - 1}\log {\left( {1 - \frac{1}{{{a_k}}}} \right)} \leq \sum\limits_{k = 1}^{n - 1}({ - \frac{1}{{{a_k}}}}) \leq \sum\limits_{k = 1}^{n - 1} \frac{-1}{k+\alpha}\to-\infty$$
Hence $a_n-n\to 0$.
4) Define $c_n=a_n-n$ and $b_n=nc_n$; we have: $$b_{n+1}-b_n=c_n(\frac{c_n}{n+c_n}-\frac{1}{n+c_n}).$$ The term on the right goes to zero and therefore $b_{n+1}-b_n\to 0$.