# limit of a sequence

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• Mar 16th 2009, 06:13 AM
xalk
limit of a sequence
if a>0 and $x_{1}>0$

and $x_{n+1} = \frac{1}{2}(x_{n} +\frac{a}{x_{n}})$

find the limit of the sequence as n goes to infinity
• Mar 16th 2009, 06:17 AM
Jhevon
Quote:

Originally Posted by xalk
if a>0 and $x_{1}>0$

and $x_{n+1} = \frac{1}{2}(x_{n} +\frac{a}{x_{n}})$

find the limit of the sequence as n goes to infinity

i assume you have already proven that a (finite) limit exists?

note that $\lim_{n \to \infty}x_n = \lim_{n \to \infty}x_{n + 1}$, call this limit $L$. Then we have

$\lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \frac 12 \left( x_n +\frac a{x_n} \right)$

$\Rightarrow L = \frac 12 \left( L +\frac aL \right)$

Now find $L$
• Mar 16th 2009, 10:14 AM
xalk
o.k thanks i have done that L= sqrt(a) or L= -sqrt(a)

Now which is the limit of the function ,because up to now we assumed that the sequence converges we have not proved that the sequence converges
• Mar 16th 2009, 10:21 AM
Plato
Quote:

Originally Posted by xalk
o.k thanks i have done that L= sqrt(a) or L= -sqrt(a)
Now which is the limit of the function ,because up to now we assumed that the sequence converges we have not proved that the sequence converges

The terms of the sequence are all positive.
So which is it?

To prove convergence, show that the sequence is monotone and bounded.
• Mar 16th 2009, 11:54 AM
xalk
Thank you, but i am not so sure why we must take the positive sqrt of a as the limit of the function because all the terms of the sequence are positive .Is there a theorem or another problem ,saying that if all the terms of the sequence are +ve then the limit of the function must be +ve??

To prove convergence i tried hard with no result so a complete proof will be very mush appreciated it
• Mar 16th 2009, 12:15 PM
Plato
It is known that if a positive sequence converges its limit is non-negative.
• Mar 16th 2009, 12:49 PM
xalk
can this be proved? if ,yes how??
• Mar 16th 2009, 01:03 PM
Plato
If the limit of a sequence is negative then almost all of its terms must be negative.
Say $\left( {s_n } \right) \to L < 0$ then using the definition let $\varepsilon = \frac{{ - L}}{2}$.
$\left( {\forall n \geqslant N} \right)\left[ {\left| {s_n - L} \right| <\varepsilon } \right]\, \Rightarrow \,s_n < \frac{L}{2} < 0$
• Mar 16th 2009, 02:15 PM
xalk
Good .So a proof of convergence is left
• Mar 16th 2009, 02:27 PM
Plato
Quote:

Originally Posted by xalk
So a proof of convergence is left

Show it is bounded and eventually monotone.
• Mar 16th 2009, 02:57 PM
xalk
HOW??
• Mar 16th 2009, 03:14 PM
Plato
Quote:

Originally Posted by xalk
HOW??

That is your job. It is so messy that I have no desire to even try it.
I am glad I don’t have to do it. Like I said, it is your problem.
• Mar 20th 2009, 05:30 PM
xalk
By considering the difference :
$x_{n}-x_{n+1} = \frac{(x_{n}-\sqrt{a})(x_{n}+\sqrt{a})}{2x_{n}}$ we can study the monotony of the sequence .

But in this difference all terms are positive except the term $x_{n}-\sqrt{a}$ which can be positive or -ve for all values of nεN.

At the same time if we can prove whether the term $x_{n}-\sqrt{a}$ is +ve or -ve we will have proved whether the sequence is bounded above or bellow by sqrt(a).

$s_n^2 = x + (s_{n-1} - xs_{n-1})^2$