# Convergence of sequence

• Apr 4th 2010, 11:07 AM
Brotha
Convergence of sequence
A) How do you prove that if 0(</=)x(</=)10, then 0(</=)sqrt(x+1)(</=)10?

B) So once that is found, then how can you prove that if 0(</=)u(</=)v(</=)10, then 0(</=)sqrt(u+1)(</=)sqrt(v+1)(</=)10?

C) They give a recursively defined sequence: a_1=0.3; a_(n+1)=sqrt((a_n)+1)for n>1
How do you find out the first five terms for it (I know what they are). then prove that this sequence converges. What is a specific theorem that will guarantee convergence, along with the algebraic results of parts A and B?

D) How do you find out the exact limit of the sequence defined in part C? Are you supposed to square the recursive equation and take limits using limit theorems? If so, then which are these theorems?
• Apr 4th 2010, 12:45 PM
atompunk
A) Graph it or use a table. Also remember that sqrt (|x|) is going to be less than |x| for |x| > 1 and if |x| < 1 than sqrt (|x|) < 1 though greater than x
• Apr 4th 2010, 02:01 PM
chisigma
The sequence can be expressed as...

$\Delta_{n} = a_{n+1} - a_{n} = f(a_{n}) = 1 + \sqrt{a_{n}} - a_{n}$ (1)

The function that generates the sequence...

$f(x) = 1 + \sqrt{x} - x$ (1)

... is represented in figure...

http://digilander.libero.it/luposabatini/MHF53.bmp

... and, because is has only one fixed point at $x_{0} = \frac{3 + \sqrt{5}}{2} = 2,6180339887\dots$ and that is an attractive fixed point, any $a_{0} \ge 0$ will produce a sequence convergent at $x_{0}$ without oscillations, because the slope of $f(x)$ in $x=x_{0}$ is in absolute value less than $1$...

Kind regards

$\chi$ $\sigma$
• Apr 4th 2010, 08:31 PM
Brotha
Quote:

Originally Posted by chisigma
The sequence can be expressed as...

$\Delta_{n} = a_{n+1} - a_{n} = f(a_{n}) = 1 + \sqrt{a_{n}} - a_{n}$ (1)

The function that generates the sequence...

$f(x) = 1 + \sqrt{x} - x$ (1)

... is represented in figure...

http://digilander.libero.it/luposabatini/MHF53.bmp

... and, because is has only one fixed point at $x_{0} = \frac{3 + \sqrt{5}}{2} = 2,6180339887\dots$ and that is an attractive fixed point, any $a_{0} \ge 0$ will produce a sequence convergent at $x_{0}$ without oscillations, because the slope of $f(x)$ in $x=x_{0}$ is in absolute value less than $1$...

Kind regards

$\chi$ $\sigma$

How did you find out delta_n?
• Apr 4th 2010, 09:46 PM
chisigma
The sequence is defined as...

$a_{n+1} = 1 + \sqrt{a_{n}}$ (1)

... so that is...

$\Delta_{n} = a_{n+1} - a_{n} = 1 + \sqrt{a_{n}} - a_{n}$ (2)

Kind regards

$\chi$ $\sigma$
• Apr 5th 2010, 05:02 AM
HallsofIvy
If $0\le x\le 10$ then $1\le x+1\le 11$ so $1\le \sqrt{x+1}\le \sqrt{10}< 10$
• Apr 5th 2010, 06:06 AM
chisigma
After a more carefull reading it seems to Me that the sequence is defined as...

$a_{n+1} = \sqrt {1 + a_{n}}$ (1)

If that is true I apologize for my mistake (Worried) ... fortunately the solving procedure is almost the same we have described. The (1) can be alternatively written as...

$\Delta_{n} = a_{n+1} - a_{n} = f(a_{n}) = \sqrt{1 + a_{n}} - a_{n}$ (2)

... the 'generating function' of which is...

$f(x)= \sqrt{1 + x} - x$ (3)

... that is represented in figure...

http://digilander.libero.it/luposabatini/MHF54.bmp

As in the previous case there is only one attractive fixed point in $x_{0} = \frac{1 + \sqrt{5}}{2} = 1,6180339887\dots$ and, since the slope of $f(x)$ at $x=x_{0}$ is in absolute value less than $1$, all $a_{0} \ge -1$ will produce a sequence converging at $x_{0}$ without oscillations...

Kind regards

$\chi$ $\sigma$
• Apr 5th 2010, 08:13 AM
Brotha
Yes the correction is true. Thank you very much.
• Apr 5th 2010, 08:15 AM
Brotha
So, what would be the exact limit of the sequence defined in part c? This is just for myself. But would you square the recursive equation and take limits using some limit theorems?
• Apr 5th 2010, 09:50 PM
chisigma
Let suppose that a sequence is defined by an initial term $a_{0}$ an the recursive relation...

$\Delta_{n} = a_{n+1} - a_{n} = f(a_{n})$ (1)

... where $f(x)$ is a continous function in $a \le x \le b$. From (1) it follows immediately that ...

$\lim_{n \rightarrow \infty} a_{n} = a_{0} + \sum_{n=0}^{\infty} \Delta_{n}$ (2)

The limit (2) will exist only if the series converges and a necessary condition for that is...

$\lim_{n \rightarrow \infty} \Delta_{n} =0$ (3)

In order to find a sufficient condition let suppose now that $f(x)$ has in $[a,b]$ a single zero in $x=x_{0}$ and that in that interval is $|f(x)|\le |r(x)|$ where $r(x)$ is a linear function crossing the x-axis in $x_{0}$ with slope equal to $-1$. Such situation is illustrated in figure...

http://digilander.libero.it/luposabatini/MHF54.bmp

... where $r(x)$ is the red line. Under these condition we can use the root test to verify the convergence of the series in (2) and we find that $\forall n$ is...

$\frac{\Delta_{n+1}}{\Delta_{n}} < 1$ (4)

... so that the series converges and for (3) the sequence converges at $x_{0}$...

Kind regards

$\chi$ $\sigma$