Question:

Step 1: Convert the sum into a Riemann integral. We recognize this sum as a Riemann sum. Let $x = \dfrac{r}{n}$, so that $\dfrac{1}{n} = dx$. As $n \to \infty$, the sum becomes: $$\lim_{n\to\infty} \sum_{r=1}^{n} \frac{\pi}{n} \cdot \frac{1}{\sin\left(\frac{(n+r)\pi}{4n}\right)} = \int_0^1 \frac{\pi}{\sin\left(\frac{(1+x)\pi}{4}\right)} dx$$ Step 2: Apply substitution to simplify the integral. Let $u = \dfrac{(1+x)\pi}{4}$. Then $du = \dfrac{\pi}{4}dx$, which gives us $dx = \dfrac{4}{\pi}du$. When $x = 0$: $u = \dfrac{\pi}{4}$ When $x = 1$: $u = \dfrac{2\pi}{4} = \dfrac{\pi}{2}$ Substituting into the integral: $$\int_0^1 \frac{\pi}{\sin\left(\frac{(1+x)\pi}{4}\right)} dx = \int_{\pi/4}^{\pi/2} \frac{\pi}{\sin u} \cdot \frac{4}{\pi} du = 4\int_{\pi/4}^{\pi/2} \csc u \, du$$ Step 3: Evaluate the antiderivative of cosecant. The antiderivative of $\csc u$ is $-\ln|\csc u + \cot u|$ or equivalently $\ln\left|\tan\dfrac{u}{2}\right|$. Using the second form: $$4\int_{\pi/4}^{\pi/2} \csc u \, du = 4\left[\ln\left|\tan\frac{u}{2}\right|\right]_{\pi/4}^{\pi/2}$$ Step 4: Apply the limits of integration. $$= 4\left(\ln\left|\tan\frac{\pi}{4}\right| - \ln\left|\tan\frac{\pi}{8}\right|\right)$$ Since $\tan\dfrac{\pi}{4} = 1$: $$= 4\left(\ln 1 - \ln\tan\frac{\pi}{8}\right) = 4\left(0 - \ln\tan\frac{\pi}{8}\right) = -4\ln\tan\frac{\pi}{8}$$ Step 5: Use the known value of $\tan\dfrac{\pi}{8}$. We know that $\tan\dfrac{\pi}{8} = \sqrt{2} - 1$. Therefore: $$-4\ln(\sqrt{2}-1) = 4\ln\frac{1}{\sqrt{2}-1}$$ Step 6: Rationalize the denominator to reach the final form. Rationalizing $\dfrac{1}{\sqrt{2}-1}$ by multiplying by $\dfrac{\sqrt{2}+1}{\sqrt{2}+1}$: $$\frac{1}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1$$ Therefore: $$4\ln\frac{1}{\sqrt{2}-1} = 4\ln(\sqrt{2}+1)$$ **Final Answer:** The value of the limit is $\boxed{4\ln(\sqrt{2}+1)}$, which corresponds to **Option 3**.