Question:

Step 1: Simplify the general term of the series. The general term in the sum is $\ln\left(\sqrt[n]{\dfrac{4r^2}{n^2}}\right)$ where $r$ ranges from $1$ to $n$. Using logarithm properties: $$\ln\left(\sqrt[n]{\dfrac{4r^2}{n^2}}\right) = \dfrac{1}{n}\ln\left(\dfrac{4r^2}{n^2}\right) = \dfrac{1}{n}\ln\left(\dfrac{2r}{n}\right)^2 = \dfrac{2}{n}\ln\left(\dfrac{2r}{n}\right)$$ Step 2: Express the entire sum in summation form. The complete sum can be written as: $$S = \sum_{r=1}^{n} \dfrac{2}{n}\ln\left(\dfrac{2r}{n}\right)$$ Step 3: Convert the Riemann sum to an integral as $n \to \infty$. As $n \to \infty$, the Riemann sum $\sum_{r=1}^{n} \dfrac{2}{n}\ln\left(\dfrac{2r}{n}\right)$ converges to the integral: $$\lim_{n \to \infty} S = 2\int_0^1 \ln(2x)\,dx$$ Step 4: Evaluate the integral using logarithm properties. Expand $\ln(2x)$ as: $$2\int_0^1 \ln(2x)\,dx = 2\int_0^1 [\ln 2 + \ln x]\,dx$$ $$= 2\left[\ln 2 \cdot x + \int \ln x\,dx\right]_0^1$$ Step 5: Apply integration by parts for $\int \ln x\,dx$. Using integration by parts with $u = \ln x$ and $dv = dx$: $$\int \ln x\,dx = x\ln x - x$$ Therefore: $$2\int_0^1 [\ln 2 + \ln x]\,dx = 2\left[\ln 2 \cdot x + x\ln x - x\right]_0^1$$ Step 6: Evaluate the definite integral. At $x = 1$: $\ln 2 \cdot 1 + 1 \cdot \ln 1 - 1 = \ln 2 + 0 - 1 = \ln 2 - 1$ At $x = 0$: The limit as $x \to 0^+$ of $x\ln x$ is $0$, so the contribution is $0$ $$= 2[(\ln 2 - 1) - 0] = 2(\ln 2 - 1) = 2\ln 2 - 2$$ Step 7: Match with the correct option. The answer is $2\ln 2 - 2$, which matches **Option 2: $2\ln(2)-2$**.