3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Think about what you know about $a_n$ (substitute) $\endgroup$ – Julian Rachman Oct 23 '15 at 7:35
2
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

Let's start with some observations:

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 $b_{n+1}-b_n\to 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.