Closed form for $\sqrt[2]{a\sqrt[3]{(a+1)\sqrt[4]{(a+2)\cdots}}}$, where $a\in\mathbb{N}$?

Question

While reviewing a document relating to nested radicals, I came across the following expression-

$$\epsilon=\sqrt[2]{a\sqrt[3]{(a+1)\sqrt[4]{(a+2)\cdots}}}$$ (where $a\in \mathbb{N})$, and thought if there would exist a closed form for it.

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My attempt-

$$\epsilon=\sqrt[2]{a\sqrt[3]{(a+1)\sqrt[4]{(a+2)\cdots}}} \tag1$$

Upon simplifyng further,

$$\frac{\epsilon^{2}}{a}={\sqrt[3]{(a+1)\sqrt[4]{(a+2)\cdots}}} \tag2$$

Upon simplifyng further,

$$\frac{\epsilon^{6}}{(a)^3(a+1)}={\sqrt[4]{(a+2)\sqrt[5]{(a+3)\cdots}}} \tag3$$

Upon simplifyng further,

$$\frac{\epsilon^{24}}{(a)^{12}(a+1)^4(a+2)}={\sqrt[5]{(a+3)\sqrt[6]{(a+4)\cdots}}} \tag4$$

Upon simplifyng further, we tend to notice the factorials of numbers like $2,3,4,...$ etcetra, which may suggest to further recurrance.

I had asked the same problem to my higher ups, and they said that 'The lower part diverges while root converges' and more details relating it to the generalisation with Ramanujan's identities.

My final doubt to be uptaken would be to find a general recurrence for this nested radical expression and any other insight on it.

Thank you.

Answer 1

Let $\epsilon_n$ be the result of truncating the expression after $n$ layers. For example:

$$\epsilon_1=\sqrt{a}$$ $$\epsilon_2=\sqrt{a\sqrt[3]{a+1}} = a^{1/2}(a+1)^{1/6}$$ $$\epsilon_3=\sqrt[2]{a\sqrt[3]{(a+1)\sqrt[4]{a+2}}} = a^{1/2}(a+1)^{1/6}(a+2)^{1/24}$$ $$\epsilon_4=\sqrt[2]{a\sqrt[3]{(a+1)\sqrt[4]{(a+2)\sqrt[5]{a+3}}}}=a^{1/2}(a+1)^{1/6}(a+2)^{1/24}(a+3)^{1/120}$$

At this point, a clear pattern emerges:

$$\boxed{\epsilon = \prod_{k=0}^{\infty} (a + k)^{\frac{1}{(k+2)!}}}$$

I don't know if there's a closed form for it, but I've found numerically that for sufficiently large values of $a$,

$$\boxed{\epsilon \approx a^{e-2}}$$

Answer 2

Let $$ \epsilon= \prod_{n=0}^\infty (a+n)^{\frac{1}{(n+2)!}} $$ Taking the logarithm $$ S(a)=\log \epsilon = \sum_{n=0}^\infty \frac{1}{(n+2)!} \log(a+n) $$ I wan't able to find a closed expression for the function, but I tryed differentiating in $a$. The derivative is $$ S'(a)= \sum_{n=0}^\infty \frac{1}{(n+2)!(a+n)} $$ Wolfram alpha can handle this and get the "closed form " (after simplifying the complex exponentials) $$ S'(a)= \frac{1}{(a-1)(a-2)} \left((e-2)a+(3-2e) + \int_0^1 u^{a-1}e^u du\right) $$ The integral in $a$ does not seems to return any result in terms of standard mathematical functions. Notice how for large $a$ the leading terms has asymptotic $$ S'(a) \approx (e-2) \frac{1}{a} $$ that leads to $$ S(a) \approx \log(a) (e-2) $$ Consistently with the numerical results found by Dan.

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If we want to prove the asymtotics directly $$ \log(a+k)= \log(a)+ \log(1+k/a) $$ $$ \log\epsilon = \log(a) (e-2) + \sum_{k=0}^\infty \frac{\log(1+k/a)}{(k+2)!}=: \log(a) (e-2) + R(a) $$ By Taylor expandind the logarithm (in the limit $a \to \infty$) $$ R(a)= \frac{1}{a} \sum_{k=0}^\infty \frac{k}{(k+2)!} + O(a^{-2}) $$ Using \begin{align} \sum_{k=0}^\infty \frac{k}{(k+2)!}&= \sum_{k=0}^\infty \frac{1}{(k+1)!}-2 \sum_{k=0}^\infty \frac{1}{(k+2)!}\\ &=(e-1)-2(e-2)=3-e \end{align} So $$ S(a)= \log(a)(e-2) +\frac{(3-e)}{a}+O(a^{-2}) $$ The same result can be obtained by integrating the asymptotic extension of the derivative