Question:

$\textcolor{green}{\textbf{Key Idea}}$ The \(k\)-th term is \[ \ln\sqrt[n]{\frac{4k^2}{n^2}} =\frac{1}{n}\ln\left(\frac{4k^2}{n^2}\right). \] Therefore the whole expression is \[ \frac{1}{n}\sum_{k=1}^{n} \ln\left(4\left(\frac{k}{n}\right)^2\right). \] This is a Riemann sum for \[ \int_0^1 \ln(4x^2)\,dx. \] $\textcolor{blue}{\textbf{Solution}}$ The terms inside the logarithms are \[ \frac{4}{n^2},\frac{16}{n^2},\frac{36}{n^2},\ldots,\frac{4n^2}{n^2}. \] The \(k\)-th term is \[ \frac{4k^2}{n^2}. \] Hence the expression is \[ \lim_{n\to\infty} \sum_{k=1}^{n} \ln\sqrt[n]{\frac{4k^2}{n^2}}. \] Using \[ \ln\sqrt[n]{A}=\frac{1}{n}\ln A, \] we get \[ \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n} \ln\left(\frac{4k^2}{n^2}\right). \] Now \[ \frac{4k^2}{n^2} =4\left(\frac{k}{n}\right)^2. \] So the limit becomes \[ \int_0^1 \ln(4x^2)\,dx. \] Now \[ \ln(4x^2)=\ln 4+2\ln x. \] Therefore \[ \int_0^1 \ln(4x^2)\,dx =\int_0^1 \ln 4\,dx+2\int_0^1 \ln x\,dx. \] Since \[ \int_0^1 \ln x\,dx=-1, \] we get \[ \int_0^1 \ln(4x^2)\,dx =\ln 4-2. \] But \[ \ln 4=2\ln 2. \] Hence the value is \[ 2\ln 2-2. \] \[ \boxed{2\ln 2-2} \] $\textcolor{red}{\textbf{Key Trap}}$ Do not forget that the logarithm is of an \(n\)-th root: \[ \ln\sqrt[n]{A}=\frac{1}{n}\ln A. \] That factor \(1/n\) is what turns the sum into a Riemann sum. Also, \[ \int_0^1 \ln x\,dx=-1, \] even though \(\ln x\) is improper at \(0\).