# Thread:

1. ## 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?

2. 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

3. 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...

... 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

$\chi$ $\sigma$

4. 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...

... 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?

5. 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

$\chi$ $\sigma$

6. If $0\le x\le 10$ then $1\le x+1\le 11$ so $1\le \sqrt{x+1}\le \sqrt{10}< 10$

7. 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 ... 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...

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

$\chi$ $\sigma$

8. Yes the correction is true. Thank you very much.

9. 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?

10. 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...

... 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

$\chi$ $\sigma$
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