How to find the value of $\sqrt{20+\sqrt{20+\sqrt{20 + \cdots}}}$

Question

Generally, I know how to calculate the square roots or cube roots, but I am confused in this question, not knowing how to do this:

$$\sqrt{20+\sqrt{20+\sqrt{20 + \cdots}}}$$

Note: Answer given in the key book is $5$. Not allowed to use a calculator.

Your question title is wrong, I think. You want the value of the expression ($5$), not its square root ($\sqrt 5$). — TonyK, Jul 30 '14 at 08:30

score 8 · Accepted Answer · edited Apr 13 '17 at 12:21

8

HINT:

Let $\displaystyle S=\sqrt{20+\sqrt{20+\sqrt{20+\cdots}}}$ which is definitely $>0$

$\displaystyle\implies S^2=20+S\iff S^2-S-20=0$

But we need to show the convergence of the sum

edited Apr 13 '17 at 12:21

Community

answered May 10 '14 at 14:53

lab bhattacharjee

See also, http://math.stackexchange.com/questions/589288/sqrt7-sqrt7-sqrt7-sqrt7-sqrt7-cdots-approximation – lab bhattacharjee May 10 '14 at 14:59
Got it, thanks! – Vivek May 10 '14 at 15:04

score 5 · Answer 2 · answered May 10 '14 at 14:54

5

Denote the corresponding value by $x$, then it satisfies the relation $$x=\sqrt{20+x},$$ with the only positive solution $x=5$.

answered May 10 '14 at 14:54

2

With the usual caveat that this only shows that if the limit exists, then it is equal to 5. (However, it seems quite easy to show that the sequence is monotone and bounded.) – Martin Sleziak Jul 30 '14 at 09:10

2 Answers2