$\displaystyle a_{n+1}=\frac{4a_{n}-3}{a_{n}}$
$\displaystyle a_{1}=2$
A.
prove that for n$\displaystyle \geq$1
1$\displaystyle \leq a_{n}\leq$3
?
B.
prove that $\displaystyle \mbox{\ensuremath{a_{n}}} converges$
and find the limit
?
$\displaystyle a_{n+1}=\frac{4a_{n}-3}{a_{n}}$
$\displaystyle a_{1}=2$
A.
prove that for n$\displaystyle \geq$1
1$\displaystyle \leq a_{n}\leq$3
?
B.
prove that $\displaystyle \mbox{\ensuremath{a_{n}}} converges$
and find the limit
?
it's a sequence not a serie.
Here are some hints:
A) Induction using the fact that the function defined by is increasing (because f'(x) is postive) then induction using
B) 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 .
We don't have to use Lagrange or Cauchy , those are simple sequences.
Let's stay with your problem:
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$\displaystyle \leq a_{n}\leq$3 and f is increasing we have
End of induction .
B) let's prove that (a_{n}) is increasing :
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 'initial value' $\displaystyle u_{0}$ is not specified and that isn't a minor detail. The 'recursive relation' can be written as...
$\displaystyle \displaystyle \Delta_{n}= u_{n+1}-u_{n}= u^{2}_{n} - \frac{u_{n}}{4}= f(u_{n})$ (1)
The function f(*) is represented here...
There is only one 'attractive fixed point' at $\displaystyle x_{0}=0$ and that means that, if the sequence converges, it converges to 0. In particular the sequence converges monotonically for $\displaystyle -\frac{3}{4} \le u_{0} < \frac{1}{4}$, converges 'with oscillation' for $\displaystyle -1 < u_{0} < -\frac{3}{4}$ and diverges for $\displaystyle u_{0}< -1$ and $\displaystyle u_{0}> \frac{1}{4}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$