# Thread: Infinite summation involving square root of 5

1. ## 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

2. 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.

3. 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 didn´t see snowtea's post.

4. 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!

5. 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.

6. 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

7. $\displaystyle \displaystyle\lim_{n \to \infty}f(n)=\sqrt{5+\sqrt{5+\sqrt{5+........}}}$

$\displaystyle \displaystyle\lim_{n \to \infty}\left[f(n)\right]^2=5+\sqrt{5+\sqrt{5+\sqrt{5+.....}}}=5+\lim_{n \to \infty}f(n)$

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

$\displaystyle \displaystyle\ a^2-a=5\Rightarrow\ a^2-a+\frac{1}{4}=\frac{21}{4}$

$\displaystyle \displaystyle\ \left[a-\frac{1}{2}\right]^2=\frac{21}{4}\Rightarrow\ a=\sqrt{\frac{21}{4}}+\frac{1}{2}$