1. **State the problem:** We want to analyze the infinite series $$\sum_{k=0}^\infty \left( \sin\left( \frac{(k+1)\pi}{12k+11} \right) - \sin\left( \frac{k\pi}{12k-1} \right) \right)$$ and find the nth partial sum $S_n$ and then evaluate the limit as $n \to \infty$. 2. **Identify the nth term:** The nth term of the series is $$a_k = \sin\left( \frac{(k+1)\pi}{12k+11} \right) - \sin\left( \frac{k\pi}{12k-1} \right).$$ 3. **Check for telescoping pattern:** Notice that the terms involve sine functions with arguments depending on $k$ and $k+1$. Let's define $$x_k = \frac{k\pi}{12k - 1}$$ and $$y_k = \frac{(k+1)\pi}{12k + 11}.$$ The term is $a_k = \sin(y_k) - \sin(x_k)$. If $y_k = x_{k+1}$, the series telescopes. 4. **Verify if $y_k = x_{k+1}$:** Calculate $x_{k+1}$: $$x_{k+1} = \frac{(k+1)\pi}{12(k+1) - 1} = \frac{(k+1)\pi}{12k + 12 - 1} = \frac{(k+1)\pi}{12k + 11} = y_k.$$ So indeed, $y_k = x_{k+1}$. 5. **Rewrite the series using $x_k$:** $$a_k = \sin(x_{k+1}) - \sin(x_k).$$ 6. **Express the partial sum $S_n$:** $$S_n = \sum_{k=0}^n a_k = \sum_{k=0}^n \left( \sin(x_{k+1}) - \sin(x_k) \right) = \sin(x_{n+1}) - \sin(x_0).$$ 7. **Calculate $x_0$ and $x_{n+1}$:** $$x_0 = \frac{0 \cdot \pi}{12 \cdot 0 - 1} = \frac{0}{-1} = 0,$$ $$x_{n+1} = \frac{(n+1)\pi}{12(n+1) - 1} = \frac{(n+1)\pi}{12n + 11}.$$ 8. **So the partial sum is:** $$S_n = \sin\left( \frac{(n+1)\pi}{12n + 11} \right) - \sin(0) = \sin\left( \frac{(n+1)\pi}{12n + 11} \right).$$ 9. **Evaluate the limit as $n \to \infty$:** $$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \sin\left( \frac{(n+1)\pi}{12n + 11} \right) = \sin\left( \lim_{n \to \infty} \frac{(n+1)\pi}{12n + 11} \right).$$ 10. **Calculate the inside limit:** $$\lim_{n \to \infty} \frac{(n+1)\pi}{12n + 11} = \lim_{n \to \infty} \frac{n+1}{12n + 11} \pi = \frac{1}{12} \pi = \frac{\pi}{12}.$$ 11. **Final limit:** $$\lim_{n \to \infty} S_n = \sin\left( \frac{\pi}{12} \right) = \sin\left( 15^\circ \right) = \frac{\sqrt{6} - \sqrt{2}}{4}.$$ **Answer:** $$S_n = \sin\left( \frac{(n+1)\pi}{12n + 11} \right), \quad \lim_{n \to \infty} S_n = \frac{\sqrt{6} - \sqrt{2}}{4}.$$