# Thread: Convergence Question

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:

$\displaystyle a=\sqrt{xa}$

so

$\displaystyle a^2-ax=0$

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

RonL

RonL

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

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

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

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

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

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