# Thread: Limit of infinite sequence.

1. ## Limit of infinite sequence.

The sequence is for x>y>0
$
a_1 = x +y,
$

$
a_n = x + y - \frac{xy}{a_{n-1}}
$

Now what i have got is
$
a_n = \frac{x^{n+1}-y^{n+1}}{x^n - y^n}
$

How do i find the limit when n goes to infinity ?

Edit: I suspect it has somthing to do with (I dont know the correct english word) the rule that when:
$
(a_n) \leq (c_n) \leq (b_n)
$

$
\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} b_n
$

Then:
$
\lim\limits_{n \to \infty} c_n = \lim\limits_{n \to \infty} a_n
$

But I am unable to construct two such limits.

2. Originally Posted by hjortur
The sequence is for x>y>0
$
a_1 = x +y,
$

$
a_n = x + y - \frac{xy}{a_{n-1}}
$

Now what i have got is
$
a_n = \frac{x^{n+1}-y^{n+1}}{x^n - y^n}
$

How do i find the limit when n goes to infinity ?

Edit: I suspect it has somthing to do with (I dont know the correct english word) the rule that when:
$
(a_n) \leq (c_n) \leq (b_n)
$

$
\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} b_n
$

Then:
$
\lim\limits_{n \to \infty} c_n = \lim\limits_{n \to \infty} a_n
$

But I am unable to construct two such limits.
Do you know that it converges? If you do, then you know $\{a_n\}\to a$, so you can just take the limit of both sides now.

$\lim_{n\to\infty}a_n = \lim_{n\to\infty}\left(x+y-\frac{xy}{a_{n-1}}\right) \implies a = x+y-\frac{xy}{a}$

Just solve for $a$ and you're done. If you don't know $a_n$ converges, I would first try to prove that it does and then use the above method.

3. Having tried out the sequence a few times with various $x$ and $y$, it seems like the sequence converges to $x$, which is a solution of the above equation.

4. I was able to find a:
$
\lim_{n\to\infty}a_n = \lim_{n\to\infty}\left(x+y-\frac{xy}{a_{n-1}}\right) \implies a = x+y-\frac{xy}{a}
$

And the solution is a=x or a=y.
The thing is I need to be able to show that it converges to x, and not y. How would I go about doing that?

5. The limit is x, not y, because x> y. The initial value is x+y> x (because both x and y are positive) and the sequence decreases. The limit is x basically because it gets to x first!

6. Jus shortly after writing this message I was able to show that the limit is x. Like i said, I knew the limit is x. But I needed to show that analyticaly.