# Math Help - series convergens mn1 q1

1. ## series convergens mn1 q1

$a_{n+1}=\frac{4a_{n}-3}{a_{n}}$
$a_{1}=2$
A.
prove that for n $\geq$1
1 $\leq a_{n}\leq$3
?
B.
prove that $\mbox{\ensuremath{a_{n}}} converges$
and find the limit
?

2. it's a sequence not a serie.
Here are some hints:
A) Induction using the fact that the function defined by $f(x)=\frac{4x-3}{x}$ is increasing (because f'(x) is postive) then induction using $f(a_{n})=a_{n+1}$
B) $(a_{n})$ is increasing and bounded (is that the word because i don't study maths in english ) so it convergences then solve f(x)=x to compute the limit.
Hard luck .

3. i only started to learn it
i tried to look on google for similar
solved questions but i dint find any.

do you have any web site which explain the theory of this subject
?
and some solved similar questions

4. First of all , i wanna know if you understood my hints , if you didn't i can write the complete solution .
I actually have some similar problems in french and arabic , wait some minutes so that i can translate and post them

5. $u_{n+1}=(u_{n})^2+\frac{3}{4}u_{n}$
we consider the function f such that $f(x)=(x)^2+\frac{3}{4}x$
1)show that $f(\left [ 0 \right;\frac{1}{4} ])\subset \left [ 0 \right;\frac{1}{4} ]$
2) show that for every positive integer n : $0\leq u_{n}\leq \frac{1}{4}$
3)Study the monotony of $(u_{n})$ then deduce that it converges.
4) compute it's limit .

Hard luck .

6. cant understand the first condition
you say that the values of F output is a part of its range
?

on convergence we have lagrange test cauchy test

you didnt say any thing about it?

7. We don't have to use Lagrange or Cauchy , those are simple sequences.
A) we will use mathematical induction:
for n=1 we have 1=<2=<3 so 1=<a_{0}=<3
we assume that it's true for n and let's prove it for n+1
since 1 $\leq a_{n}\leq$3 and f is increasing we have $1=f(1)\leq f(a_{n})\leq f(3)=3\Leftrightarrow 1\leq a_{n+1}\leq 3$
End of induction .
B) let's prove that (a_{n}) is increasing :
$a_{n+1}-a_{n}=\frac{4a_{n}-3-\left (a_{n} \right )^2}{a_{n}}=\frac{-(a_{n}-3)(a_{n}-1)}{a_{n}}\geq 0$ we have now proved that it's increasing .
Now and since it's bounded , it converges.
As i said , to compute the limit , solve the equation f(x)=x

$u_{n+1}=(u_{n})^2+\frac{3}{4}u_{n}$
we consider the function f such that $f(x)=(x)^2+\frac{3}{4}x$
1)show that $f(\left [ 0 \right;\frac{1}{4} ])\subset \left [ 0 \right;\frac{1}{4} ]$
2) show that for every positive integer n : $0\leq u_{n}\leq \frac{1}{4}$
3)Study the monotony of $(u_{n})$ then deduce that it converges.
4) compute it's limit .

Hard luck .
The 'initial value' $u_{0}$ is not specified and that isn't a minor detail. The 'recursive relation' can be written as...

$\displaystyle \Delta_{n}= u_{n+1}-u_{n}= u^{2}_{n} - \frac{u_{n}}{4}= f(u_{n})$ (1)

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

There is only one 'attractive fixed point' at $x_{0}=0$ and that means that, if the sequence converges, it converges to 0. In particular the sequence converges monotonically for $-\frac{3}{4} \le u_{0} < \frac{1}{4}$, converges 'with oscillation' for $-1 < u_{0} < -\frac{3}{4}$ and diverges for $u_{0}< -1$ and $u_{0}> \frac{1}{4}$...

Kind regards

$\chi$ $\sigma$

9. Actually u_0=1/5 .
Sorry i was in a hurry.

We don't have to use Lagrange or Cauchy , those are simple sequences.
A) we will use mathematical induction:
for n=1 we have 1=<2=<3 so 1=<a_{0}=<3
we assume that it's true for n and let's prove it for n+1
since 1 $\leq a_{n}\leq$3 and f is increasing we have $1=f(1)\leq f(a_{n})\leq f(3)=3\Leftrightarrow 1\leq a_{n+1}\leq 3$
End of induction .
B) let's prove that (a_{n}) is increasing :
$a_{n+1}-a_{n}=\frac{4a_{n}-3-\left (a_{n} \right )^2}{a_{n}}=\frac{-(a_{n}-3)(a_{n}-1)}{a_{n}}\geq 0$ we have now proved that it's increasing .
Now and since it's bounded , it converges.
As i said , to compute the limit , solve the equation f(x)=x
We don't have to use Lagrange or Cauchy , those are simple sequences.
A) we will use mathematical induction:
for n=1 we have 1=<2=<3 so 1=<a_{0}=<3
we assume that it's true for n and let's prove it for n+1
since 1 $\leq a_{n}\leq$3 and f is increasing we have $1=f(1)\leq f(a_{n})\leq f(3)=3\Leftrightarrow 1\leq a_{n+1}\leq 3$
End of induction .
because a_n is between 3 and 1 the numerator is negative
and denominator is positive so the whole thing is negative

not positive
B) let's prove that (a_{n}) is increasing :
$a_{n+1}-a_{n}=\frac{4a_{n}-3-\left (a_{n} \right )^2}{a_{n}}=\frac{-(a_{n}-3)(a_{n}-1)}{a_{n}}\geq 0$ we have now proved that it's increasing .
Now and since it's bounded , it converges.
As i said , to compute the limit , solve the equation f(x)=x
We don't have to use Lagrange or Cauchy , those are simple sequences.
A) we will use mathematical induction:
for n=1 we have 1=<2=<3 so 1=<a_{0}=<3
we assume that it's true for n and let's prove it for n+1
since 1 $\leq a_{n}\leq$3 and f is increasing we have $1=f(1)\leq f(a_{n})\leq f(3)=3\Leftrightarrow 1\leq a_{n+1}\leq 3$
End of induction .
because a_n is between 3 and 1 the numerator is negative
and denominator is positive so the whole thing is negative

not positive
B) let's prove that (a_{n}) is increasing :
$a_{n+1}-a_{n}=\frac{4a_{n}-3-\left (a_{n} \right )^2}{a_{n}}=\frac{-(a_{n}-3)(a_{n}-1)}{a_{n}}\geq 0$ we have now proved that it's increasing .
Now and since it's bounded , it converges.
As i said , to compute the limit , solve the equation f(x)=x
the numerator is negative
denominator is positive so the whole thing is negative
not possitive

11. so to find the limit we put L instead of a_n and a_n+1 and compute the roots?