Cool Summation!! (Hint: Fundamental something something)

Relevant wiki: Riemann Sums

$\begin{aligned} L & = \lim_{n \to \infty} \left[\ln \left(\sqrt [n]{\frac 4{n^2}}\right) + \ln \left(\sqrt [n]{\frac {16}{n^2}}\right) + \ln \left(\sqrt [n]{\frac {36}{n^2}}\right) + \cdots + \ln \left(\sqrt [n]{\frac {4n^2}{n^2}}\right) \right] \\ & = \lim_{n \to \infty} \sum_{k=1}^n \ln \left(\sqrt [n]{\frac {4k^2}{n^2}}\right) \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac 1n \ln \left(\frac {4k^2}{n^2}\right) & \small \color{#3D99F6} \text{By Riemann sums: } \\ & = \int_0^1 \ln (4x^2)\ dx & \small \color{#3D99F6} \lim_{n \to \infty} \sum_{k=a}^b \frac 1n f\left(\frac kn\right) = \lim_{k \to n} \int_\frac ak^\frac bk f(x) \ dx \\ & = \int_0^1 \left(2\ln 2 + 2 \ln x\right)\ dx \\ & = 2x\ln 2 + 2x\ln x -2x\ \bigg|_0^1 \\ & = 2\ln 2 + 2\ln 1 - 2x - 0 - 2 {\color{#3D99F6} \lim_{x \to 0} x \ln x} - 0 & \small \color{#3D99F6} \text{See note: }\lim_{x \to 0} x \ln x = 0 \\ & = \boxed{2\ln 2 - 2} \end{aligned}$

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Note:

$\begin{aligned} L & = \lim_{x \to 0} x\ln x \\ & = \lim_{x \to 0} \frac {\ln x}{\frac 1x} & \small \color{#3D99F6} \text{A }\infty/\infty \text{ cases, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {\frac 1x}{-\frac 1{x^2}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t }x. \\ & = \lim_{x \to 0} -x \\ & = 0 \end{aligned}$

I think you can use Stirling's approximation you get

$\begin{aligned} L&=\lim_{n\to\infty}\sum\limits_{k=1}^n\ln\left(\sqrt[n]{\frac{ 4k^2}{n^2}}\right)=\lim_{n\to\infty}\ln\left(\sqrt[n]{\frac{\prod\limits_{k=1}^n 4k^2}{n^{2n}}}\right)=\lim_{n\to\infty}\ln\left(\frac{4}{n^2}\sqrt[n]{\prod\limits_{k=1}^n k^2}\right)\\ &=\lim_{n\to\infty}\ln\left(\frac{4}{n^2}\sqrt[n]{n!n!}\right)\\ &=\lim_{n\to\infty}\ln\left(\frac{4}{n^2}\frac{n^2}{e^2}\sqrt[n]{2\pi n}\right)=\ln\left(\frac{4}{e^2}\cdot 1\cdot 1\right)\\ &=2\ln(2)-2 \end{aligned}$

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edit: If you think, it is too inexact because we use a approximation: No, because $n\to\infty$ :

More exactly you have: $n!=\sqrt{2\pi n}\left(\frac n{\mathrm e}\right)^n\mathrm e^{\mu(n)}$ with $0<\mu(x)<1/(12x)\,,\,x>0~~\Rightarrow \lim\limits_{n\to\infty}\mu(n)=0$ , and so:

$\begin{aligned} L_2&=\lim_{n\to\infty}\ln\left(\sqrt[n]{n!n!}\right)=\lim_{n\to\infty}\ln\left(\frac{n^2}{e^2}\sqrt[n]{2\pi n}\right)+\lim_{n\to\infty}\ln\left(\mathrm e^{2\mu/n}\right)\\ &=\lim_{n\to\infty}\ln\left(\frac{n^2}{e^2}\sqrt[n]{2\pi n}\right) \end{aligned}$

Rearranging terms and using the log identity ln(a)+ln(b)+...ln(z) = ln(ab...z) gives us $\lim_{n \to \infty} (\frac{2}{n} \ln(\frac{2^n n!}{n^n}))$ Using Stirling's approximation for n! ( $\sqrt{2\pi}\ n^{n+{\small\frac12}}e^{-n} \le n! \le e\ n^{n+{\small\frac12}}e^{-n}$ ), we get $\lim_{n \to \infty} 2 \ln\big(\frac{2}{e} \sqrt[n]{k\sqrt{n}}\big)$ where $\sqrt{2\pi}\ \le k \le e\$ . This gives us the desired result $2 \ln2-2$

Amateur explanation:

This can be rewritten as a definite integral with bounds 0 to 2 as shown:

$\lim_{n\to\infty} \sum_{i=1}^{n}ln((\frac{2i}{n})^{(\frac{2}{n})})$

$=\lim_{n\to\infty} \sum_{i=1}^{n}(\frac{2}{n})ln(\frac{2i}{n})$

$=\lim_{n\to\infty}(\frac{2}{n}) \sum_{i=1}^{n}ln(\frac{2i}{n})$

$=\int_{0}^{2} ln(x) dx$

$=2ln(2)-2$