Worked exercise — Nested radical

Problem

Verify that the nested radical with $n$ levels $a_n = \sqrt{2 + \sqrt{2 + \sqrt{2+\cdots}}}$ satisfies the recurrence $a_{n+1} = \sqrt{2 + a_n}$ with $a_0 = \sqrt{2}$; show that it is increasing and bounded by $2$, and compute its limit.

Solution

Recurrence. Adding a level to the radical means nesting the previous expression under a new root: $a_{n+1} = \sqrt{2 + a_n}$, with $a_0 = \sqrt2 \approx 1{,}414$.

Boundedness ($a_n < 2$), by induction. We have $a_0 = \sqrt2 < 2$. If $a_n < 2$, then $a_{n+1} = \sqrt{2+a_n} < \sqrt{2+2} = 2$. Hence $a_n < 2$ for every $n$.

Increasing. We have $a_{n+1} > a_n \iff \sqrt{2+a_n} > a_n \iff 2 + a_n > a_n^2 \iff a_n^2 - a_n - 2 < 0 \iff (a_n-2)(a_n+1) < 0$, true for $-1 < a_n < 2$, a condition that is satisfied. The sequence is therefore increasing.

Limit. Increasing and bounded above $\Rightarrow$ it converges to a value $\ell$. Passing to the limit in the recurrence $\ell = \sqrt{2+\ell}$, that is $\ell^2 - \ell - 2 = 0$, whence $(\ell-2)(\ell+1) = 0$. Discarding the negative root: $\ell = \boxed{2}.$

Links

Topics: Sequences
Concepts: Limit of a sequence · Monotonicity · Bounded sequence · Recursively defined sequence
Skills: Computing limits · Proving
Exercise type: Computing a limit · Proof