1. Convergence Question

Consider the following sequence

with a1 = and x>1.Show that it converges and that its limit is x.
hint:Start by using induction to show that this sequence is non-decreasing and bounded above by x.

2. Originally Posted by matty888
Consider the following sequence

with a1 = and x>1.Show that it converges and that its limit is x.
hint:Start by using induction to show that this sequence is non-decreasing and bounded above by x.
Take the hint, that shows the sequence converges. Then the limit satisfies the equation:

$a=\sqrt{xa}$

so

$a^2-ax=0$

hence $a=0$ or $a=x$. But the first of these is imposible, so $a=x$.

RonL

RonL

3. Note that:
• $a_{n}=x^{\frac{1}{2}}\cdot{a_{n-1}^{\frac{1}{2}}}$
• $a_{n}=x^{\frac{1}{2}}\cdot{\left(x^{\frac{1}{2}}\c dot{a_{n-2}^{\frac{1}{2}}}\right)^{\frac{1}{2}}}$ $=x^{\frac{1}{2}+\frac{1}{4}}\cdot{a_{n-2}^{\frac{1}{4}}}$
• $a_n=x^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}\cdot{a _{n-3}^{\frac{1}{8}}}$
And so on.

The general formula is: $
a_n = x^{\sum\nolimits_{j = 1}^k {\left( {\tfrac{1}
{2}} \right)^j } } \cdot a_{n - k} ^{{\raise0.5ex\hbox{\scriptstyle 1}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{\scriptstyle {2^k }}}}
$
for $
1 \leqslant k \leqslant n-1
$

We now let $k=n-1$ ==> $
a_n = x^{\sum\nolimits_{j = 1}^{n - 1} {\left( {\tfrac{1}
{2}} \right)^j } } \cdot a_1 ^{\tfrac{1}
{{2^{n - 1} }}}

$

Thus: $
\mathop {\lim }\limits_{n \to + \infty } a_n = \mathop {\lim }\limits_{n \to + \infty } x^{\sum\nolimits_{j = 1}^{n - 1} {\left( {\tfrac{1}
{2}} \right)^j } } \cdot a_1 ^{\tfrac{1}
{{2^{n - 1} }}}

$

Applying the Geometric Sum formula we get $
a_n = x^{\sum\nolimits_{j = 1}^{n - 1} {\left( {\tfrac{1}
{2}} \right)^j } } \cdot a_1 ^{\tfrac{1}
{{2^{n - 1} }}} = x^{1 - \left( {\tfrac{1}
{2}} \right)^n }

$

Therefore: $
\mathop {\lim }\limits_{n \to + \infty } a_n = x
$