Compute $\lim_{n\to\infty}{a_n}$

Question

Given $a$ and $b$ such that $a \ge b > 0,$ define the sequence ${a_n}$ by $$a_1 = a+b , \ \ \ a_n= a_1 - \frac {ab}{a_{n-1}} , \ \ \ \ \ n \ge 2.$$

Determine the $n^{\rm th}$ term of the sequence and compute $\lim_{n\to\infty}{a_n}$

I take $l = l - \frac {ab}{l}$ after that $ l= l - \frac{ab}{l}$, $l^2 =l^2-ab$, and I get $ab = 0$.

I don't know how I can find $\lim_{n\to\infty}{a_n}$.

Answer 1

Assuming the limit exists, you would have (since $a_1 = a+b$) $$ l = a_1 - \frac{ab}{l} \implies l^2-(a+b) l +ab =0 $$ so you get that the limit is $l=a$.

Answer 2

The limit, assuming it exists, will satisfy $$ L=a+b-\frac{ab}L. $$ This has roots $a$ and $b$, so this is not enough.

When $a=b$, we get with ease that $a_n=a(1+\frac1n)$, and so the limit is $a$.

So now assume that $a>b$. With a little playing with the formulas, one can guess that $$\tag1 a_n=\frac{a^{n+1}-b^{n+1}}{a^n-b^n}. $$ Indeed, if $a_n$ is as above, \begin{align} a_{n+1}&=a+b-\frac{ab}{\frac{a^{n+1}-b^{n+1}}{a^n-b^n}} =a+b-\frac{ab(a^n-b^n)}{a^{n+1}-b^{n+1}}\\ \ \\ &=\frac{(a+b)(a^{n+1}-b^{n+1})-ba^{n+1}+ab^{n+1}}{a^{n+1}-b^{n+1}}\\ \ \\ &=\frac{a^{n+2}-b^{n+2}}{a^{n+1}-b^{n+1}},\\ \ \\ \end{align} which implies by induction that $a_n$ is as in $(1)$ for all $n$. Now, looking at $(1)$, and since $a> b\geq0$, $$ a_n=a\,\frac{1-\left(\frac ba\right)^{n+1}}{1-\left(\frac ba\right)^{n}}\xrightarrow[n\to\infty]{}a. $$