Two nested radicals

From A Nested Radical we get $\sqrt[4]{5\sqrt[7]{5^2\sqrt[10]{5^3\sqrt[13]{5^4\sqrt[16]{5^5\sqrt[19]{\cdots}}}}}}=\sqrt[3]{5} \approx 1.709976$ .

$\sqrt[3]{5\sqrt[6]{5\sqrt[9]{5\sqrt[12]{5\sqrt[15]{5\sqrt[18]{\cdots}}}}}}=\left(5 \left(5 \left(5 \left(\cdots \right)^\frac 1{12} \right)^\frac 19 \right)^\frac 16 \right)^\frac 13=\exp \left(\ln 5 {\color{#3D99F6} \left(\frac 13 + \frac 13 \cdot \frac 16 + \frac 13 \cdot \frac 16 \cdot \frac 19 + \cdots \right)} \right) \small \color{#3D99F6} \text{where }\exp (x) = e^x. \\ \implies \sqrt[3]{5\sqrt[6]{5\sqrt[9]{5\sqrt[12]{5\sqrt[15]{5\sqrt[18]{\cdots}}}}}}=\exp \left(\ln {5^{\sqrt[3]{e}-1}} \right) =5^{\sqrt[3]{e}}-1 \approx 1.8902586 \text{( See \# below)} \\ \therefore \text{the answer is } 1.8902586-1.709976 \approx \color{#3D99F6}0.18$

#Proof of $\displaystyle \color{#3D99F6} \left(\frac 13 + \frac 13 \cdot \frac 16 + \frac 13 \cdot \frac 16 \cdot \frac 19 + \cdots \right)=\sqrt[3]{e}-1$

We know, $n!=n(n-1)(n-2)(n-3)\cdots 2\cdot 1$ hence,

$3^n\cdot n!=(3n)(3n-3)(3n-6)(3n-9)\cdots 6\cdot 3$ writing in reverse we get $3^n\cdot n!=3\cdot 6\cdot 9\cdots (3n-9)(3n-6)(3n-3)(3n)$

Also, $\displaystyle e^x-1=\sum_{n=1}^{\infty}{\frac {x^n}{n!}}$

Putting $\displaystyle x=\frac 13$ we get:

$\displaystyle \sqrt[3]{e}-1=\sum_{n=1}^{\infty}{\frac 1{3^n\cdot n!}}=\left(\frac 13 + \frac 13 \cdot \frac 16 + \frac 13 \cdot \frac 16 \cdot \frac 19 + \cdots \right)$

$\begin{aligned} P_1 & = \sqrt[3]{5\sqrt[6]{5\sqrt[9]{5\sqrt[12]{\cdots}}}} = \left(5 \left(5 \left(5 \left(5 \cdots \right)^\frac 1{12} \right)^\frac 19 \right)^\frac 16\right)^\frac 13 = 5^\frac 13 5^{\frac 13 \cdot \frac 16}5^{\frac 13 \cdot \frac 16 \cdot \frac 19} \cdots = \prod_{k=1}^\infty 5^{\frac 1{3^kk!}} \\ & = \exp \left(\ln 5 \sum_{k=1}^\infty \frac 1{3^k k!}\right) = \exp \left( \ln 5 (e^\frac 13-1)\right) \approx 1.8903 \end{aligned}$

$\begin{aligned} P_2 & = \sqrt[4]{5\sqrt[7]{5^2\sqrt[10]{5^3\sqrt[13]{\cdots}}}} & \small \color{#3D99F6} \text{Since }\sqrt[x+1]{n\sqrt[2x+1]{n^2\sqrt[3x+1]{n^3\sqrt[4x+1]{\cdots}}}} = \sqrt[x]n \text{ (see reference)} \\ & = \sqrt[3] 5 \approx 1.7010 \end{aligned}$

$\implies P_1 - P_2 \approx \boxed{0.18}$ .

---

Reference: A Nested Radical