<p>Let \(S_n = c - \dfrac{1}{(n+1)(n+2)(n+3) \cdot 3}\). Find \(\lim_{n \to \infty} S_n\) (given that \(c = \dfrac{1}{18}\)).</p>
Step-by-Step Solution
Key Concept: Recognize that the fraction term vanishes as n→∞, leaving only the constant c. The denominator (n+1)(n+2)(n+3)·3 grows without bound, making the fraction approach zero.
<p><strong>Step 1:</strong> Analyze the limit as n→∞:</p><p>$$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left(c - \frac{1}{(n+1)(n+2)(n+3) \cdot 3}\right)$$</p><p><strong>Step 2:</strong> Evaluate the behavior of each term. As n→∞, the denominator (n+1)(n+2)(n+3)·3 grows to infinity (cubic growth), so the fractional term approaches 0:</p><p>$$\lim_{n \to \infty} \frac{1}{(n+1)(n+2)(n+3) \cdot 3} = 0$$</p><p><strong>Step 3:</strong> Apply the limit:</p><p>$$\lim_{n \to \infty} S_n = c - 0 = c = \frac{1}{18}$$</p><p>∴ Answer: <strong>1/18</strong></p>
Correct Answer: 19