Nested-Radical Sequence Limits

Repeated Square Roots as a Sequence

A nested radical has several layers of square roots, so its appearance can make it difficult to tell what happens as the number of layers increases. The key is to describe each finite radical as one term of a sequence. Then the new outer square root is produced from the previous term by a simple recurrence.

For example, the finite expressions $\sqrt{1}$, $\sqrt{1+\sqrt{1}}$, and $\sqrt{1+\sqrt{1+\sqrt{1}}}$ are successive terms of a sequence. Each new term is obtained by taking the square root of $1$ plus the preceding term.

Definition: Fix $a>0$ and a starting value $x_0\ge 0$. Define the sequence recursively by $$ x_{n+1}=\sqrt{a+x_n},\qquad n\ge 0. $$ Its terms are finite nested radicals: $x_1=\sqrt{a+x_0}$, $x_2=\sqrt{a+\sqrt{a+x_0}}$, and so on. The usual nested radical with the same constant at every level starts with $x_0=0$, so $x_1=\sqrt a$.

The condition $a>0$ and the starting condition $x_0\ge 0$ ensure the recurrence is defined at every step. Indeed, if $x_n\ge 0$, then $a+x_n>0$, so $x_{n+1}$ is a real, nonnegative number. Thus, by induction, every term is real and nonnegative.

Find a Fixed-Point Candidate

If the sequence converges to a limit $L$, the recurrence suggests what equation that limit must satisfy. The function $t\mapsto\sqrt{a+t}$ is continuous on the relevant domain. Therefore, by the limit law for continuous functions of convergent sequences, taking limits in the recurrence gives $L=\sqrt{a+L}$. This is a necessary equation for a limit, but solving it alone does not prove that the sequence converges.

L=\sqrt{a+L} \quad\Longrightarrow\quad L^2-L-a=0.

The quadratic equation has two real roots when $a>0$: $L=(1+\sqrt{1+4a})/2$ and $L=(1-\sqrt{1+4a})/2$. The second root is negative because $\sqrt{1+4a}>1$. But the output of a real square root is nonnegative, so a limit of the sequence must be nonnegative. The only possible fixed-point candidate is therefore $L=(1+\sqrt{1+4a})/2$.

Domain check: For $a>0$, the equation $L=\sqrt{a+L}$ requires $L\ge 0$. Its unique nonnegative solution is $$ L=\frac{1+\sqrt{1+4a}}{2}. $$ It satisfies the original equation because $L^2=L+a$ and $L\ge0$, so $\sqrt{a+L}=\sqrt{L^2}=L$.

Why the Recurrence Converges

A fixed point is only a candidate: a sequence can fail to converge even when its recurrence has a fixed point. Here we can prove convergence directly by estimating the distance between each term and the candidate $L$. The identity for the difference of two square roots provides the needed estimate.

Compare one term with the fixed point.
Since $x_{n+1}=\sqrt{a+x_n}$ and $L=\sqrt{a+L}$, factor the difference of their squares to obtain an exact error identity.

Bound the denominator away from zero.
Every $x_n\ge0$, and $a>0$, so the denominator in the identity is at least $\sqrt a+L$, a number greater than $1$.

Iterate the error bound.
The distance to $L$ is multiplied by at most a fixed number $q<1$ at each step. The resulting geometric bound tends to zero.

Here is the complete argument. From $x_{n+1}^2=a+x_n$ and $L^2=a+L$, subtract to get $x_{n+1}^2-L^2=x_n-L$. Factoring the left-hand side gives the error identity below. Its denominator is positive because both square roots are nonnegative and $L>0$.

|x_{n+1}-L| = \frac{|x_n-L|}{x_{n+1}+L} = \frac{|x_n-L|}{\sqrt{a+x_n}+L}.

Since $x_n\ge0$, we have $\sqrt{a+x_n}\ge\sqrt a$. Also, $L>1$ for $a>0$, so $\sqrt a+L>1$. Set $q=1/(\sqrt a+L)$; then $0<q<1$. The identity gives the estimate $|x_{n+1}-L|\le q|x_n-L|$. Applying this estimate repeatedly yields $|x_n-L|\le q^n|x_0-L|$ for every $n\ge0$. The geometric sequence $q^n$ tends to zero when $0<q<1$, so the error tends to zero. This proves that $x_n\to L$.

Theorem: If $a>0$ and $x_0\ge0$, the recurrence $x_{n+1}=\sqrt{a+x_n}$ is defined for every $n$, and it converges to $$ \frac{1+\sqrt{1+4a}}{2}. $$ More precisely, with $q=1/(\sqrt a+L)$ and $L=(1+\sqrt{1+4a})/2$, it satisfies $|x_n-L|\le q^n|x_0-L|$.

Worked Examples

Worked Example: The Classic Radical with Ones

Consider the finite nested radicals defined by $x_0=0$ and $x_{n+1}=\sqrt{1+x_n}$. The first terms are $x_1=\sqrt{1+0}=1$, $x_2=\sqrt{1+1}=\sqrt2$, and $x_3=\sqrt{1+\sqrt2}$.

A limit $L$ must satisfy $L=\sqrt{1+L}$. Squaring gives $L^2=1+L$, or $L^2-L-1=0$. The roots are $(1+\sqrt5)/2$ and $(1-\sqrt5)/2$. Since $L$ must be nonnegative, keep only $L=(1+\sqrt5)/2$. Check it in the original equation: $L^2=L+1$, and $L>0$, so $\sqrt{1+L}=\sqrt{L^2}=L$.

For this sequence, $q=1/(1+L)$, which lies strictly between $0$ and $1$. The error estimate proves $|x_n-L|\le q^n|0-L|=q^nL\to0$. Thus the finite radicals converge to $(1+\sqrt5)/2$, not merely to a value suggested by the fixed-point equation.

Worked Example: A Radical with Limit 2

Set $x_0=0$ and $x_{n+1}=\sqrt{2+x_n}$. The first terms are $x_1=\sqrt2$, $x_2=\sqrt{2+\sqrt2}$, and $x_3=\sqrt{2+\sqrt{2+\sqrt2}}$. A limit must solve $L=\sqrt{2+L}$, so $L^2-L-2=0$, or $(L-2)(L+1)=0$.

The roots are $2$ and $-1$. Only $2$ is nonnegative. Substitution checks the candidate: $\sqrt{2+2}=\sqrt4=2$. For convergence, the theorem applies because $a=2>0$ and $x_0=0\ge0$. Specifically, $q=1/(\sqrt2+2)<1$, and $|x_n-2|\le q^n|0-2|=2q^n\to0$. Therefore this nested radical converges to $2$.

Worked Example: Changing the Innermost Starting Value

Let $x_0=1$ and $x_{n+1}=\sqrt{3+x_n}$. The first terms are $x_1=\sqrt{3+1}=2$, $x_2=\sqrt{3+2}=\sqrt5$, and $x_3=\sqrt{3+\sqrt5}$. This sequence corresponds to finite nested radicals with a different innermost value from the usual choice $x_0=0$.

The fixed-point equation is $L=\sqrt{3+L}$, hence $L^2-L-3=0$. Its roots are $(1+\sqrt{13})/2$ and $(1-\sqrt{13})/2$. The first is positive and the second is negative, so only $L=(1+\sqrt{13})/2$ is possible. For the check, $L^2=L+3$, and positivity gives $\sqrt{3+L}=\sqrt{L^2}=L$.

Since $a=3>0$ and $x_0=1\ge0$, the convergence estimate applies with $q=1/(\sqrt3+L)<1$. It gives $|x_n-L|\le q^n|1-L|\to0$. Changing the starting value changes the finite terms, but in this example it does not change the limit.

A Common Fixed-Point Mistake

Solving the quadratic and choosing either root without checking the original square-root equation can produce an impossible answer. For the radical with $a=1$, one might solve $L^2-L-1=0$ and select $L=(1-\sqrt5)/2$, approximately $-0.618$, simply because it is a root of the quadratic. But a real square root cannot equal a negative number. Substituting this value into the unsquared equation makes the failure explicit: $\sqrt{1+L}$ is positive, whereas $L$ is negative.

Correction: Squaring can introduce algebraic candidates that do not satisfy the original equation. Check the domain and substitute the candidate back into $L=\sqrt{a+L}$. For $a>0$, only the positive root $(1+\sqrt{1+4a})/2$ passes that check.

Why the Error Estimate Matters

The fixed-point equation identifies where a convergent sequence would have to end up. The error estimate supplies the separate argument that the terms actually approach that point. In this setting, each iteration reduces the error by at least a fixed proportion: the estimate $|x_{n+1}-L|\le q|x_n-L|$, with $q<1$, is stronger than simply knowing that the fixed point exists.

The bound also clarifies the role of the assumptions. Nonnegative starting values keep every square root real and ensure $\sqrt{a+x_n}\ge\sqrt a$. The positive constant $a$ gives a fixed point greater than $1$, which makes $q<1$. Without checking the domain and the size of the error factor, solving the quadratic alone would not establish convergence.

For standard nested radicals, the method is therefore systematic: write the recurrence, solve for its nonnegative fixed point, and prove convergence by bounding the difference from that point. The recurrence and error estimate turn a visual pattern of deeper and deeper radicals into a precise limit argument.

Check Your Understanding

For each question, distinguish the fixed-point calculation from the convergence argument.

Write the first three terms for $x_0=0$, $x_{n+1}=\sqrt{4+x_n}$, and find its nonnegative fixed-point candidate.
Why must a limit of a real sequence defined by $x_{n+1}=\sqrt{a+x_n}$ be nonnegative?
For $a=1$, explain why the negative root of $L^2-L-1=0$ is not a valid limit.
What conditions on $a$ and $x_0$ ensure that the stated convergence estimate applies?
In the error estimate, why is $q=1/(\sqrt a+L)$ strictly less than $1$ when $a>0$?

What You'll Learn