1. prove that

x^n/n! = 0 for n-> 0
using sandwich theorem

a sequence given by a(n+1) = SQRT(2 + an)
prove that it is converging.. I found the limit is 2.. but how to prove it converges?

prove that if an converges to a and bn converges to b then an/bn converges to a/b using the theorem |an-l| < E for all n > N and for any E > 0

2. Re: prove that

Originally Posted by ice_syncer
... a sequence given by a(n+1) = SQRT(2 + an)... prove that it is converging.. I found the limit is 2.. but how to prove it converges?...
The recursive relation can be written as...

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

The function f(x) is represented here...

There is only one 'attractive fixed point' [i.e. a point where f(x) crosses the x axis with negative slope...] at $x_{0}=2$ and, because for $x>-2$ is $|f(x)|<|x_{0}-x|$ [see the 'red line'...] any $a_{0}>-2$ will produce a sequence converging to 2 monotonically...

Kind regards

$\chi$ $\sigma$

3. Re: prove that

How can you show it without graph ?

4. Re: prove that

If in general we write the 'recursive relation' as...

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

... then is...

$a_{n}= a_{0}+ \sum_{k=0}^{n-1} \Delta_{k}$ (2)

... and the convergence of the series in (2) implies the convergence of the $a_{n}$. If we suppose that $f(x)$ is continous and vor all n is $\Delta_{n}>0$ or $\Delta_{n}<0$, then the series in (2) converges if...

a) $\lim_{n \rightarrow \infty} \Delta_{n}=0$...

b) for n 'large enough' is $\frac{\Delta_{n+1}}{\Delta_{n}} \le \lambda < 1$ [ratio test]...

The condition a) is satisfied if it exist a real number $x_{0}$ for which is $f(x_{0}=0)$, and in such case is [if it exists...] $\lim_{n \rightarrow \infty} a_{n}= x_{0}$. The condition b) is satisfied if is $f^{'}(x_{0}) <0$ and there exist an interval around $x_{0}$ where is $|f(x)|< |x_{0}-x|$...

Kind regards

$\chi$ $\sigma$

5. Re: prove that

lim(n->inf)(x^n/n!)
case 1: |x|<=1
=> 0<=|x^n/n!|<=|1/n!| for all n>=1 => lim(n->inf)(x^n/n!)=0 by the squeeze thm for all |x|<1

case 2: |x|>1 ;
let c(x)=Product(for i from 1 to ceiling(|x|) of x/i) is constant w.r.t n
let d(x)=|x|/(ceiling(|x|)+1) =>0<d(x)<1
lim(n->inf)(d(x)^n)=0, |x/i|<=d(x) for all i>ceiling(|x|)

0<=|x^n/n!|=|c(x)||Product(for i from (ceiling(|x|)+1) to n of x/i)|
<=|c(x)|(Product for i from (ceiling(|x|)+1) to n of d(x))
=|c(x)|d(x)^n. for all natural numbers n

hence by the squeeze thm x^n/n! ->0 for |x|>1

Thus for all real numbers x, x^n/n! ->0 for n->inf

The main idea i think is to write the nth term as a product over i of x/i and to consider the first cieling(x) terms as a constant. each consective term is x/n times the previous so if it were to converge to L, L=lim(n->inf)(x^n/n!)=lim(n->inf)(x/n)*lim(n->inf)(x^(n-1)/(n-1)!)=0*L
so L must be 0.

6. Re: prove that

Originally Posted by ice_syncer
prove that if an converges to a and bn converges to b then an/bn converges to a/b using the theorem |an-l| < E for all n > N and for any E > 0
first note: b must be non zero for a/b to exist so we require b nonzero
second note: b is nonzero and b_n->b => S={k:b_k=0} is finite so let n_0=max(S).
we begin by deleting the first n_0 terms from both the sequences {a_n} and {b_n} (this will not affect convergence as we do not alter the 'tails' of the sequences) and adjusting the indices so they both start at 1. This is in order to ensure {a_n/b_n} is a sequence of real numbers. From now on I assume b_n is non zero for all n.

b_n converges to a real number b and a is a real number
=> there exists max{(b_n)}, max{a/(b*b_n)} in real numbers

let c=max{(b_n)}, d=max{a/(b*b_n)}

i) {a_n}->a => for all E>0 there exists N s.t. |a_n-a|<E*c/2 for all n>=N
ii) {b_n}->b => for all E>0 there exists M s.t. |b_n-b|<E*d/2 for all n>=M

0<=|a_n/b_n-a/b|
=|(b*a_n)-(a_n*b)|/|b_n*b|
=|(b*a_n-b*a)+(b*a-b_n*a)|/|b_n*b|
<=(|(b*a_n-b*a)|+|(b*a-b_n*a)|)/|b_n*b|
=(|a_n-a|/|b_n|)+(|a/(b*b_n)||b-b_n|)
<=E/2+E/2=E for n>= max(M,N)

hence a_n/b_n->a/b if n->inf

7. Re: prove that

Originally Posted by ice_syncer
x^n/n! = 0 for n-> 0
using sandwich theorem
Let $a=\lceil |x|\rceil$

$|x^n|\leq a^n=a^aa^{n-a}\leq a^a\frac{(n-1)!}{(a-1)!}$

(The last inequality holds because there are $n-a$ terms in $\frac{(n-1)!}{(a-1)!}$, all of which are $\geq a$.)

Given this,

$0\leq\left|\frac{x^n}{n!}\right|\leq\frac{a^a}{(a-1)!}\frac{(n-1)!}{n!}=\frac{a^a}{(a-1)!}\cdot\frac{1}{n}$

Hence the squeeze theorem says the limit is 0.

An easier way would be to use the ratio test:

$\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{(n+1)! }}{\frac{x^n}{n!}}\right|=\lim_{n\to\infty}\left| \frac{x}{n+1}\right|=0$

Therefore the sum converges so the terms must go to zero.

8. Re: prove that

Isn't it suffiencient to show that it is increasing and has a limit? How do you show that the recursive sequence is increasing ?

9. Re: prove that

Originally Posted by ice_syncer
Isn't it suffiencient to show that it is increasing and has a limit? How do you show that the recursive sequence is increasing ?
There are a few cases to consider:

1. Suppose $a_n > 2$. Then $\sqrt{a_n+2} > \sqrt{2+2}=2$. Furthermore, since both sides are positive, $a_n > \sqrt{a_n+2}$ iff $a_n^2>a_n+2$ iff $a_n^2-a_2-2=(a_n-2)(a_n+1)>0$. Since $a_n > 2$ by assumption, the sequence is monotone decreasing and bounded below by $2$. So if $a_0 > 2$, the sequence is monotone decreasing and converges to $2$.

2. Suppose $0\leq a_n < 2$. Then $\sqrt{a_n+2} < \sqrt{2+2}=2$. The same reasoning as before shows that if $0\leq a_0 < 2$, the sequence is monotone increasing and converges to $2$.

3. Suppose $-2\leq a_n<0$. Then $\sqrt{a_n+2} < \sqrt{2+2}=2$ as before, and $a_n < \sqrt{a_n+2}$ because the LHS is negative and the RHS is positive, so the sequence is monotone increasing.

4. If $a_0 < -2$, this isn't a real sequence anymore, so we may ignore this case.

The conclusion is that the sequence converges to $2$ in the following manner:
$-2 \leq a_0 < 2\implies\mbox{monotone~increasing}$
$a_0=2\implies\mbox{stationary}$
$a_0 > 2\implies\mbox{monotone~decreasing}$