Jtdylktuy

Let $$a_1=3quad , quad a_{n+1}=frac{3a_n}{4}+frac{1}{a_n}$$
Show that $a_n$ converges.

I know that I need to prove that $a_n$ is monotonic and bounded.

I've assumed that the limit exists, made the calculation and get that $L=2$, hence claiming two claims:

1. 
   $a_nge 2$ (I know that 0 is a simpler bound)


2. 
   $a_n$ is monotonically decreasing.

Now, if I can prove the first claim. then I can write $$a_{n+1}=frac{3a_n}{4}+frac{1}{a_n}=frac{3a_n}{4}+frac{a_n}{a_n^2}le frac{3a_n}{4}+frac{a_n}{4}=a_n$$ which proves the second claim.

My only problem is proving the first one. If it is not the right direction, please hint me.

To show boundedness from below, you may use AM-GM:
$$frac{3a_n}{4}+frac{1}{a_n}geq 2 sqrt{frac{3a_n}{4}cdot frac{1}{a_n}}= sqrt{3}$$
For convergence you may proceed as follows:

• 
  $f(x) = frac{3x}{4} + frac{1}{x}$ has a fixpoint for $x^{star} = 2$
  


• $f'(x) = frac{3}{4}-frac{1}{x^2}$


• A quick calculation shows that $|f'(x)| < 1$ for $x > frac{2}{sqrt{7}} Rightarrow |f'(x)| leq color{blue}{q} < 1$ for $x geq sqrt{3}$

So, for any starting value $a_0 > 0$ you get
$$|2-a_{n+1}| = |f'(xi_n)||2-a_n| < color{blue}{q} |2-a_n| < color{blue}{q^n}|2-a_1|stackrel{nto infty}{longrightarrow}0 $$

Edit after comment:

Specifically for your question concerning $a_n geq 2$:
$$ color{blue}{a_{n+1}-2} = frac{3a_n}{4}+frac{1}{a_n} - left(frac{3cdot 2}{4}+frac{1}{2} right)= frac{3}{4}left( a_n - 2 right) - frac{a_n - 2}{2a_n}$$
$$ = left(frac{3}{4} - frac{1}{2a_n} right)(a_n -2) stackrel{color{blue}{a_n geq 2}}{geq} frac{1}{2}(a_n -2) color{blue}{geq 0}$$

The inequality $ x geq 2$ implies $frac {3x} 4+frac 1 x geq 2$ because$frac {3x} 4+frac 1 x$ is increasing on $x >frac 2 {sqrt3}$ (and $frac {3times 2} 4+frac 1 2 =2$). This gives (by induction) $a_n geq 2$ for all $n$.

Let $a_n=2+d_n .$ Then $$d_{n+1}=3(2+d_n)/4+ 1/(2+d_n)-2.$$ Use some elementary algebra on this to show that $d_n>0implies d_n>d_{n+1}>0.$

So by induction on $n,$ if $0<a_1-2=d_1$ then $a_n$ decreases strictly to a limit $Lgeq 2.$