VNMO 2013

[Phlembac Adib Hasan]

Define a sequence $\{a_n\}$ as: $\left\{\begin{aligned}& a_1=1 \\ & a_{n+1}=3-\frac{a_{n}+2}{2^{a_{n}}}\ \ \text{for} \ n\geq 1.\end{aligned}\right.$

Prove that this sequence has a finite limit as $n\to+\infty$ . Also determine the limit.

[zadid xcalibured]

Define the function $f(x)=\frac{x+2}{2^x}$
$f'(x)=2^{-x}$ $(1-(x+2) ln2)$.
As $x>1$,$f'(x)<2^{-x}(1-ln8)<0$
so this function is decreasing.
Which implies the sequence $a_n$ is increasing.
Now by induction we can show that $a_n \leq 2$ for all $n$.
actually this is the limit.

[Phlembac Adib Hasan]

My Proof:
I approached in the same way of Zadid bhai to show $a_i$ is non-decreasing.
Suppose for some integer $k\ge 1$, $a_{k+1}>2$. Define $g(x)=2^x-x-2$. $g(1)<0,\; \; g(2)=0$
Show that $g$ is an increasing function. So
\[a_{k+1}>2\Longleftrightarrow g(a_k)>0\Longleftrightarrow a_k>2\]
By the same argument (or better to say some kind of induction) we deduce that $a_1>2$. A contradiction. So $a_n\leq 2$. The rest is trivial.

[*Mahi*]

Define the function $f(x)=\frac{x+2}{2^x}$
$f'(x)=2^{-x}$ $(1-(x+2) ln2)$.
As $x>1$,$f'(x)<2^{-x}(1-ln8)<0$
so this function is decreasing.

Easier way to prove this function is decreasing : just take $\lg$, $\lg f(x) = \lg (x+2) - x $, which is decreasing.