# Infinite summation involving square root of 5

• Jan 10th 2011, 09:43 AM
Pranas
Infinite summation involving square root of 5
Hello.

I have this recurring function or whatever you call it:
$\displaystyle $\begin{array}{l} f(0) = 0\\ f(n) = \sqrt {5 + f(n - 1)} \end{array}$$

Once $\displaystyle n$ increases, we have $\displaystyle $\sqrt 5$$, inside of which there is another $\displaystyle $\sqrt 5$$ added and so forth (this clearly converges).

As the title says, I am looking forward to obtaining $\displaystyle $\mathop {\lim }\limits_{n \to \infty } f(n)$$

P.S. I started from the greatest depth and noticed $\displaystyle $\sqrt {5 + \sqrt 5 } = \frac{{\sqrt {20 + 4\sqrt 5 } }}{2} = \frac{{2 + \sqrt 5 }}{2}$$, however later on I am not able to control the expression...

P.P.S. Messing around I've got a good feeling that such infinite summations $\displaystyle $\sqrt {a + \sqrt {a + \sqrt {a + ...\sqrt a } } } = \frac{{1 + \sqrt {1 + 4a} }}{2}$$ can be defined like that, no clue how to prove it though
• Jan 10th 2011, 09:52 AM
snowtea
If you can show that it converges, then:

Take the limit of both sides of the recurrence.
Use the fact that $\displaystyle {\lim }\limits_{n \to \infty } f(n) = {\lim }\limits_{n \to \infty } f(n-1) = L$
to get
$\displaystyle L = \sqrt {5 + L}$

Solve for L.
• Jan 10th 2011, 09:54 AM
FernandoRevilla
Quote:

Originally Posted by Pranas
$\displaystyle f(n) = \sqrt {5 + f(n - 1)}$

If

$\displaystyle l=\displaystyle\lim_{n \to \infty}{f(n)}$

then taking limits in both menbers,

$\displaystyle l=\sqrt {5 + l}$

and necessarily the limit is the positive solution.

Fernando Revilla

Edited: Sorry, I didnīt see snowtea's post.
• Jan 10th 2011, 09:59 AM
Pranas
This gives $\displaystyle $\mathop {\lim }\limits_{n \to \infty } f(n) = \frac{{1 + \sqrt {21} }}{2}$$ almost instantly.

Smooth as silk, thanks for quick help!
• Jan 10th 2011, 10:06 AM
DrSteve
Just be aware that the above solution only works if the limit exists. To show that the limit exists, just show that the sequence is bounded above and increasing.
• Jan 10th 2011, 10:34 AM
FernandoRevilla
It is esay to prove that $\displaystyle a_n=f(n)\geq 0$ is increasing. To prove that is bounded we can use

$\displaystyle a_{n+1}^2\geq a_n^2 \Leftrightarrow\ldots \Leftrightarrow (l_1-a_n)(a_n-l_2)\geq 0$

where $\displaystyle l_1$ is the positive root of $\displaystyle l^2-l-5$ and $\displaystyle l_2$ the negative one.

Fernando Revilla
• Jan 10th 2011, 10:56 AM
$\displaystyle \displaystyle\lim_{n \to \infty}f(n)=\sqrt{5+\sqrt{5+\sqrt{5+........}}}$
$\displaystyle \displaystyle\lim_{n \to \infty}\left[f(n)\right]^2=5+\sqrt{5+\sqrt{5+\sqrt{5+.....}}}=5+\lim_{n \to \infty}f(n)$
$\displaystyle \displaystyle\lim_{n \to \infty}f(n)=a$
$\displaystyle \displaystyle\ a^2-a=5\Rightarrow\ a^2-a+\frac{1}{4}=\frac{21}{4}$
$\displaystyle \displaystyle\ \left[a-\frac{1}{2}\right]^2=\frac{21}{4}\Rightarrow\ a=\sqrt{\frac{21}{4}}+\frac{1}{2}$