This summer I set out to create a set of interesting integrals involving infinite nested radicals. In so doing, I went on to derive a number of closed formulae for infinite nested radicals with periodic base and root sequences. This section is thus devoted to describing what I’ve learned through this exercise.

#### Nested Radicals as Infinite Products

Let $(a_n)_{n\in \mathbb{Z}_{\geq 1}}$ be a periodic sequence of positive terms with period $T\in \mathbb{Z}_{\geq 1}$ so that $a_{n+T}=a_n$ for all $n\geq 1$. Let us consider the infinite nested radical

$\displaystyle l=\sqrt{a_1\sqrt{a_2\sqrt{a_3\sqrt{a_4\sqrt{\cdots}}}}}.$

Provided $l$ converges, we find from periodicity of $(a_n)$ that

$\displaystyle l=\sqrt{a_1\sqrt{a_2\sqrt{\cdots\sqrt{a_T\sqrt{a_{T+1}\sqrt{\cdots}}}}}}=\sqrt{a_1\sqrt{a_2\sqrt{\cdots\sqrt{a_T\sqrt{a_{1}\sqrt{a_2\sqrt{\cdots}}}}}}}=\sqrt{a_1\sqrt{a_2\sqrt{\cdots\sqrt{a_Tl}}}}=\prod_{n=1}^{T}a_n^{2^{-n}}\times l^{2^{-T}}$

so that rearranging gives

$\displaystyle l=\prod_{n=1}^Ta_n^{\frac{2^{-n}}{1-2^{-T}}}.$

But observe that the expression for $l$ is formally equivalent to

$\displaystyle \prod_{n=1}^{\infty}a_n^{2^{-n}}.$

Thus, we find our sequence $(a_n)$ satisfies

$\displaystyle \prod_{n=1}^{\infty}a_n^{2^{-n}}=\prod_{n=1}^Ta_n^{\frac{2^{-n}}{1-2^{-T}}}.$

Where the convergence of $l$ is concerned, this is guaranteed by the following result whenever the sum $\sum_{n=1}^{\infty}2^{-n}\log(a_n)$ converges (see Infinite Product on Wolfram MathWorld).

Theorem 1. Given a sequence of positive numbers $(b_n)_{n\geq 1}$, the product

$\displaystyle \prod_{n=1}^{\infty}b_n$

converges to a nonzero number if and only if the following series converges:

$\displaystyle \sum_{n=1}^{\infty}\log(b_n).$

Now, notice that any periodic sequence of real numbers is necessarily bounded. In particular, if $(a_n)$ is a periodic sequence of positive numbers with period $T$, we see that $M:=\max_{n\geq 1}a_n=\max_{n\in\{1,\cdots,T\}}a_n$, so that $M$ is always finite. Consequently, so is the quantity $C:=\max_{n\geq 1}|\log(a_n)|$. But then,

$\displaystyle \sum_{n=1}^{\infty}\left|2^{-n}\log(a_n)\right|\leq C\sum_{n=1}^{\infty}2^{-n}=C.$

Hence, the sum $\sum_{n=1}^{\infty}2^{-n}\log(a_n)$ converges absolutely and, as such, the product $\prod_{n=1}^{\infty}a_n^{2^{-n}}$ always converges by Theorem 1. With the formal equivalence

$\displaystyle \sqrt{a_1\sqrt{a_2\sqrt{a_3\sqrt{a_4\sqrt{\cdots}}}}}\equiv\prod_{n=1}^{\infty}a_n^{2^{-n}}$

in mind, we conclude the following.

Corollary 1. Let $(a_n)_{n\in\mathbb{Z}_{\geq 1}}$ be a periodic sequence of positive numbers with period $T\in \mathbb{Z}_{\geq 1}$ so that $a_{n}=a_{n+T}$ for all $n\geq 1$. Then

$\displaystyle \sqrt{a_1\sqrt{a_2\sqrt{a_3\sqrt{a_4\sqrt{\cdots}}}}}$

converges and there holds

$\displaystyle \sqrt{a_1\sqrt{a_2\sqrt{a_3\sqrt{a_4\sqrt{\cdots}}}}}=\prod_{n=1}^Ta_n^{\frac{2^{-n}}{1-2^{-T}}}.$

Example 1. Consider the sequence $(a_n)$ given by $a_n=a^{n}$ and $a_n=a_{n+T}$ for $n\geq 1$ with some $a>1$ and $T\in\mathbb{Z}_{\geq 1}$ given. Corollary 1 allows us to explicitly determine $l$ given a period $T\in \mathbb{Z}_{\geq 1}$. Let us write $l=l(T)$ to indicate the dependence of $l$ on $T$. We then have

$\displaystyle l(T)=a^{\frac{1}{1-2^{-T}}\sum_{n=1}^{T}n\cdot 2^{-n}}$

from which we calculate

$\displaystyle l(T)=a^{\frac{2-(T+1)\cdot2^{1-T}+T\cdot 2^{-T}}{1-2^{-T}}}.$

Notably, in the infinite period limit as $T\to \infty$ we find

$\displaystyle \lim_{T\to \infty}\sqrt{a_1\sqrt{a_2\sqrt{a_3\sqrt{a_4\sqrt{\cdots}}}}}=a^2.$

Alternatively, the same result can be deduced directly by considering $(a_n)$ without imposing any periodicity. As such we find

$\displaystyle \sqrt{a\sqrt{a^2\sqrt{a^3\sqrt{a^4\sqrt{\cdots}}}}}=\prod_{n=1}^{\infty}a^{n2^{-n}}=a^{\sum_{n=1}^{\infty}n2^{-n}}$

which gives the same result after evaluating the infinite sum in the exponent. Note that in general,

$\displaystyle \sum_{n=1}^{T}nx^{-n}=\frac{x+Tx^{-T}-(T+1)x^{1-T}}{(1-x)^2}$

if $x\neq 1$ and $T\geq 1$.

$\displaystyle \sqrt{a_1\sqrt{a_2\sqrt{a_3\sqrt{a_4\sqrt{\cdots}}}}}$

leads us to consider the same calculation for the general nested radical

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}$

where $(a_n)$ and $(b_n)$ are periodic sequences of positive numbers in-phase, which means there exists some $T\in\mathbb{Z}_{\geq 1}$ such that $a_n=a_{n+T}$ and $b_n=b_{n+T}$ for all $n\geq 1$. Let’s say that the sequence $(b_n)$ is the root sequence and $(a_n)$ is the base sequence in the above expression. In this case, we have a generalisation of Corollary 1 whose assumptions allow us to deduce immediately that the sum

$\displaystyle \sum_{n=1}^{\infty}\log(a_n)\prod_{j=1}^nb_j^{-1}$

converges absolutely. Our approach to proving Corollary 1 thus applies similarly, giving

Theorem 2. Let $(a_n)_{n\in\mathbb{Z}_{\geq 1}}$ and $(b_n)_{n\in\mathbb{Z}{\geq 1}}$ be periodic sequences of positive numbers in phase with period $T\in \mathbb{Z}_{\geq 1}$. Suppose $b_1,b_2,\cdots, b_T> 1$. Then

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}$

converges. Moreover, if $T\geq 2$ then

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^Ta_n^{\frac{b_{n+1}\cdots b_T}{b_1\cdots b_T-1}},$

while if $T=1$ then

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=a_1^{\frac{1}{b_1-1}}.$

Thus far, our analysis concerned periodic base sequences that are in phase with periodic root sequences. What can be said when

• Either sequence $(a_n)$, $(b_n)$ is not periodic, or
• Both sequences $(a_n)$ and $(b_n)$ are periodic but out of phase?

The following example considers an extreme case where $(b_n)$ isn’t periodic while $(a_n)$ is periodic.

Example 2. Suppose $(a_n)$ is a periodic sequence with period $T\in \mathbb{Z}_{\geq 1}$ so that $a_n=a_{n+T}$ for $n\geq 1$, and let $(b_n)$ be the sequence given by $b_n=n+1$, $n\geq 1$. Then, we have the expression

$\displaystyle \sqrt[2]{a_1\sqrt[3]{a_2\sqrt[4]{a_3\sqrt[5]{a_4\sqrt[6]{\cdots}}}}}\equiv\prod_{n=1}^{\infty}a_n^{\frac{1}{(n+1)!}}$

converges because the sum $\sum_{n=1}^{\infty}\frac{\log(a_n)}{(n+1)!}$ converges absolutely by periodicity of $(a_n)$. We then find

$\displaystyle \sqrt[2]{a_1\sqrt[3]{a_2\sqrt[4]{a_3\sqrt[5]{a_4\sqrt[6]{\cdots}}}}}=\prod_{n=1}^{T}a_n^{g(n+1,T)}$

where

$\displaystyle g(k,T)=\sum_{m=0}^{\infty}\frac{1}{(Tm+k)!}.$

There appears to be no closed form for $g(k,T)$ while it is clear that the sum converges. With help from Wolfram MathWorld, we may express $g$ in terms of the generalized hypergeometric function $_pF_q(c_1,\cdots, c_p; b_1,\cdots, b_q; z)$ whereby

$\displaystyle g(k,T)={}_1F_q\left(1;\frac{k+1}{T},\frac{k+2}{T},\cdots, \frac{k+T-1}{T},\frac{k}{T}+1;T^{-T}\right).$

In the case where $T=1$, so that $(a_n)$ is a constant sequence, we find

$\displaystyle \sqrt[2]{a_1\sqrt[3]{a_1\sqrt[4]{a_1\sqrt[5]{a_1\sqrt[6]{\cdots}}}}}=a_1^{e-2}.$

On the other hand, we have an example of a nested radical with periodic root sequence $(b_n)$ and non-periodic base sequence $(a_n)$.

Example 3. Take $a_n=n+1$, $b_n=2$ for $n\geq 1$. We then consider the expression

$\displaystyle \sqrt{2\sqrt{3\sqrt{4\sqrt{5\sqrt{6 \cdots}}}}}\equiv \prod_{k=1}^{\infty}(k+1)^{2^{-k}}.$

By Theorem 1 we see that the above infinite product converges because the following sum converges (by the comparison test):

$\displaystyle \sum_{k=1}^{\infty}2^{-k}\log(k+1).$

Then, with help from Wolfram Alpha, we find that

$\displaystyle \sqrt{2\sqrt{3\sqrt{4\sqrt{5\sqrt{6 \cdots}}}}}=\exp\left(-\text{\emph{LerchPhi}}^{(0,1,0)}\left(\frac{1}{2},0,1\right)\right)\approx 2.761206842$

Remark 1. We see from our two extreme examples that familiarity with special functions can be beneficial to arriving at closed forms for out-of-phase periodic nested infinite radicals.

Now, suppose $(a_n)$ and $(b_n)$ are periodic base and root sequences with respective periods $T_a$ and $T_b$, whereby $a_n=a_{n+T_a}$ and $b_n=b_{n+T_b}$ for $n\geq 1$. If $T_a=T_b+p$, for some $p\in\mathbb{Z}_{\geq 1}$ we have the following result that allows for exact computation of the corresponding nested infinite radical.

Theorem 3. Let $(a_n)_{n\in\mathbb{Z}_{\geq 1}}$ and $(b_n)_{n\in\mathbb{Z}_{\geq 1}}$ be periodic sequences of real numbers with periods $T_a,T_b\in \mathbb{Z}_{\geq 1}$ such that $a_n>0$, $a_n=a_{n+T_a}$ and $b_n=b_{n+T_b}$ for all $n\geq 1$ while $T_a=T_b+p$ for some $p\in\mathbb{Z}_{\geq 0}$. If $|b_1|,|b_2|,\cdots |b_{T_b-1}|,|b_{T_b}|>1$, then

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{T_a}a_n^{S(n)}$

where

$\displaystyle S(n)=\frac{(b_1\cdots b_{T_b})^{T_a}}{(b_1\cdots b_{T_b})^{T_a}-1}\sum_{k=0}^{T_b-1}\prod_{j=1}^{T_ak+n}b_{j}^{-1}$

for $n\in\mathbb{Z}_{\geq 1}.$

Remark 2. Notice from the conditions of Theorem 3 that we can assign to each $b_n$ an arbitrary sign, as long as $|b_n|>1$ for all $1\leq n\leq T_b$. With this flexibility, we’ve managed to cover other types of infinite nested radicals such as

$\displaystyle \sqrt[b_1]{\frac{a_1}{\sqrt[b_2]{\frac{a_2}{\sqrt[b_3]{\frac{a_3}{\sqrt[b_4]{\frac{a_4}{\sqrt[b_5]{\cdots}}}}}}}}},$

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{\frac{a_3}{\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}},$

$\displaystyle\sqrt[b_1]{a_1\sqrt[b_2]{\frac{a_2}{\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}},$

and

$\displaystyle \sqrt[b_1]{\frac{a_1}{\sqrt[b_2]{a_2\sqrt[b_3]{\frac{a_3}{\sqrt[b_4]{a_4\sqrt[b_5]{\frac{a_5}{\sqrt[b_6]{\cdots}}}}}}}}}.$

When the $T_b$ divides $T_a$, we have the following theorem whose result generalises Theorem 2.

Theorem 4. Let $(a_n)_{n\in\mathbb{Z}_{\geq 1}}$ and $(b_n)_{n\in\mathbb{Z}_{\geq 1}}$ be periodic sequences of real numbers with periods $T_a,T_b\in \mathbb{Z}_{\geq 1}$ which satisfy $T_a=mT_b$ for some $m\in\mathbb{Z}_{\geq 1}$ while $a_n>0$, $a_n=a_{n+T_a}$ and $b_n=b_{n+T_b}$ for all $n\geq 1$. If $|b_1|,|b_2|,\cdots |b_{T_b-1}|,|b_{T_b}|>1$ then

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{T_a}a_n^{S(n)}$

where

$\displaystyle S(n)=\frac{(b_1\cdots b_{T_b})^{m}}{(b_1\cdots b_{T_b})^m-1}\prod_{j=1}^{n}b_j^{-1}$

for $n\in\mathbb{Z}_{\geq 1}.$

From Theorem 3 we can deduce the case where $T_a$ is an affine function of $T_b$, in that $T_a=mT_b+p$ for some $m,p\in\mathbb{Z}$ with $m\geq 1$ and $p\geq 0$. As such, the following result generalises Theorem 4.

Theorem 5. Let $(a_n)_{n\in\mathbb{Z}_{\geq 1}}$ and $(b_n)_{n\in\mathbb{Z}_{\geq 1}}$ be periodic sequences of real numbers with periods $T_a,T_b\in \mathbb{Z}_{\geq 1}$ such that $a_n>0$, $a_n=a_{n+T_a}$ and $b_n=b_{n+T_b}$ for all $n\geq 1$, while
$T_a=mT_b+p$ for some $p,m\in\mathbb{Z}$ where $p\geq 0$ and $m\geq 1$.
Suppose $|b_1|,|b_2|,\cdots |b_{T_b-1}|,|b_{T_b}|>1$. Then

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{T_a}a_n^{S(n)}$

where

$\displaystyle S(n)=\frac{(b_1\cdots b_{T_b})^{mT_a}}{(b_1\cdots b_{T_b})^{mT_a}-1}\sum_{k=1}^{mT_b-1}\prod_{q=1}^{T_ak+n}b_q^{-1}$

for $n\in\mathbb{Z}_{\geq 1}.$

Theorem 5 covers the evaluation of our nested radicals when the base and root periods are related via $T_a=mT_b+p$. So, what can be said when the periods are swapped, i.e when $T_b=mT_a+p$? It turns out that all of our previous results are contained in the following in which no particular relationship between $T_a$ and $T_b$ needs to be assumed.

Theorem 6. Let $(a_n)_{n\in\mathbb{Z}_{\geq 1}}$ and $(b_n)_{n\in\mathbb{Z}_{\geq 1}}$ be periodic sequences of real numbers with periods $T_a,T_b\in \mathbb{Z}_{\geq 1}$ such that $a_n>0$, $a_n=a_{n+T_a}$ and $b_n=b_{n+T_b}$ for all $n\geq 1$. If $|b_1|,|b_2|,\cdots |b_{T_b-1}|,|b_{T_b}|>1$. Then

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{T_a}a_n^{S(n)}$

where

$\displaystyle S(n)=\frac{(b_1\cdots b_{T_b})^{A}}{(b_1\cdots b_{T_b})^{A}-1}\sum_{k=0}^{B-1}\prod_{j=1}^{T_ak+n}b_j^{-1}$

for $n\in\mathbb{Z}_{\geq 1}$ with $A=\frac{l.c.m(T_a,T_b)}{T_b}$ and $B=\frac{l.c.m(T_a,T_b)}{T_a}$.

Theorem 6 allows us to determine some nested radicals with periodic sequences having arbitrary periods. In particular, if we hold $T_a$ fixed, sending $T_b$ to infinity appropriately gives a formula for calculating our general nested radical whenever the root sequence $(b_n)$

• isn’t necessarily periodic,
• satisfies $|b_n|\geq 1$ for all $n\geq 1$, and
• contains a subsequence $(b_{n_j})$ for which $|b_{n_j}|\geq c$ for all $j\geq 1$, with some $c>1$, and the sequence $\left(n_{j+1}-n_{j}\right)_{j\geq 1}$ has polynomial growth.

Assuming this, consider a periodic approximation of $(b_n)$ with period $T_b$. Then, we essentially find that $|b_1\cdots b_{T_b}|\to \infty$, $B\to \infty$ and $A\to 1$ in the limit as $T_b$ is sent to infinity in such a way that $T_a$ divides $T_b$ (since then $l.c.m(T_a,T_b)=T_b$ in this limit). Consequently, we have for each $n\geq 1$,

$\displaystyle S(n)\to G(n):= \sum_{k=0}^{\infty}\prod_{j=1}^{T_ak+n}b_j^{-1}$

which implies that

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{T_a}a_n^{G(n)}$

with

$\displaystyle G(n)= \sum_{k=0}^{\infty}\prod_{j=1}^{T_ak+n}b_j^{-1}$

for $n\geq 1$. Note that the polynomial growth assumption guarantees the series $G$ converges absolutely for each $n\geq 1$. To summarise, we have

Corollary 2. Let $(a_n)_{n\in\mathbb{Z}{\geq 1}}$ be a periodic sequence of positive numbers with period $T_a$ so that $a_n=a_{n+T_a}$ for $n\geq 1$. Suppose $(b_n)_{n\in\mathbb{Z}_{\geq 1}}$ is a sequence of real numbers for which $|b_n|\geq 1$ for all $n\geq 1$ and it contains a subsequence $(b_{n_j})$ which satisfies $|b_{n_j}|\geq c$ for all $j\geq 1$, with some $c>1$, and the sequence $(n_{j+1}-n_j)_{j\geq 1}$ has polynomial growth. Then

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{T_a}a_n^{G(n)}$

where the sum

$\displaystyle G(n)=\sum_{k=0}^{\infty}\prod_{j=1}^{T_ak+n}b_j^{-1}$

converges for each $n\in\mathbb{Z}_{\geq 1}$.

As another corollary of Theorem 6 we derive a result similar to Theorem 4 but with the condition $T_b=mT_a$ instead of $T_a=mT_b$.

Corollary 3. Let $(a_n)_{n\in\mathbb{Z}_{\geq 1}}$ and $(b_n)_{n\in\mathbb{Z}_{\geq 1}}$ be periodic sequences of real numbers with periods $T_a,T_b\in \mathbb{Z}_{\geq 1}$ such that $a_n>0$, $a_n=a_{n+T_a}$ and $b_n=b_{n+T_b}$ for all $n\geq 1$ and $T_b=mT_a$ for some $m\in\mathbb{Z}_{\geq 1}$. Suppose $|b_1|,|b_2|,\cdots |b_{T_b-1}|,|b_{T_b}|>1$. Then

$\displaystyle \sqrt[b_1]{a_1\sqrt[b_2]{a_2\sqrt[b_3]{a_3\sqrt[b_4]{a_4\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{T_a}a_n^{S(n)}$

where

$\displaystyle S(n)=\frac{b_1\cdots b_{T_b}}{b_1\cdots b_{T_b}-1}\sum_{k=0}^{m-1}\prod_{j=1}^{T_ak+n}b_j^{-1}$

for $n\in\mathbb{Z}_{\geq 1}$.

Remark 3. Notice the interesting difference between the formulae of $S(n)$ in Theorem 4 and Corollary 3.

• Theorem 4: $\displaystyle S(n)=\frac{(b_1\cdots b_{T_b})^{m}}{(b_1\cdots b_{T_b})^m-1}\prod_{j=1}^{n}b_j^{-1}$
• Corollary 3: $\displaystyle S(n)=\frac{b_1\cdots b_{T_b}}{b_1\cdots b_{T_b}-1}\sum_{k=0}^{m-1}\prod_{j=1}^{T_ak+n}b_j^{-1}.$

By comparing the two formulae, we see order 1 dependence on the multiplier $m$ as $m$ tends to infinity, although the role played by this multiplier is different between the cases. In fact, taking $m$ to infinity we find

• Theorem 4: $T_a\to \infty$ while $T_b$ finite and $\displaystyle \lim_{m\to \infty}S(n)=\prod_{j=1}^{n}b_j^{-1}$
• Corollary 3: $T_b\to\infty$ while $T_a$ finite and $\displaystyle \lim_{m\to\infty}S(n)=\sum_{k=0}^{\infty}\prod_{j=1}^{T_ak+n}b_j^{-1}$

Example 4. With the formula given by Theorem 6, we can derive a curious formula for the determinant of the matrix $A=\left(a_{ij}\right)_{1\leq i,j\leq p}$ with entries given by $a_{ij}=c_i^{jS(i)}$, for some positive numbers $\left(c_i\right)_{1\leq i\leq p}$, and $S$ as given in the theorem. In particular, set $T_a=p$ and, given a permutation $\sigma \in S_p$, let our periodic base sequence $(a_i)$ be given by $a_i=c_i^{\sigma (i)}$ for $1\leq i\leq n$, leaving the root sequence $(b_n)$ unspecified (though satisfying the assumptions of Theorem 6). Then, we have

$\displaystyle \sqrt[b_1]{c_1^{\sigma(1)}\sqrt[b_2]{c_2^{\sigma(2)}\sqrt[b_3]{c_3^{\sigma(3)}\sqrt[b_4]{c_4^{\sigma(4)}\sqrt[b_5]{\cdots}}}}}=\prod_{n=1}^{p}c_n^{S(n)\sigma(n)}=\prod_{n=1}^{p}a_{n,\sigma(n)}.$

The definition of determinant of $A$ thus gives

$\displaystyle \text{det}(A)=\sum_{\sigma\in S_p}\text{sgn}(\sigma)\sqrt[b_1]{c_1^{\sigma(1)}\sqrt[b_2]{c_2^{\sigma(2)}\sqrt[b_3]{c_3^{\sigma(3)}\sqrt[b_4]{c_4^{\sigma(4)}\sqrt[b_5]{\cdots}}}}}.$

We have thus concluded our analysis of infinite nested radicals that are given by infinite products. Indeed, we have not covered explicit evaluation of all such nested radicals. As such, the reader is now invited to answer the following question.

Open Question. Are there closed forms for the following expressions?

• $\displaystyle \sqrt{2\sqrt[3]{3\sqrt[4]{4\sqrt[5]{5\sqrt[6]{\cdots}}}}}$
• $\displaystyle \sqrt{3\sqrt[3]{4\sqrt[4]{5\sqrt[5]{6\sqrt[6]{\cdots}}}}}$
• $\displaystyle \sqrt{2\sqrt[3]{2\sqrt[4]{3\sqrt[5]{3\sqrt[6]{4\sqrt[7]{4\sqrt[8]{5\sqrt[9]{5\sqrt[10]{6\sqrt[11]{\cdots}}}}}}}}}}$