Key Concepts and Formulas

• Summation of Powers:
  • $\sum_{r=1}^k r = \frac{k(k+1)}{2}$
  • $\sum_{r=1}^k r^2 = \frac{k(k+1)(2k+1)}{6}$
• Factorial Properties:
  • $n! = n \cdot (n-1)!$
  • $n! = n \cdot (n-1) \cdot (n-2)!$
• Taylor Series Expansion of $e^x$: The series for $e^x$ centered at 0 is $\sum_{k=0}^\infty \frac{x^k}{k!}$. For $x=1$, this gives $e = \sum_{k=0}^\infty \frac{1}{k!}$.

Step-by-Step Solution

Step 1: Express $S_n$ in Summation Notation and Derive its Closed Form. The given sum is $S_n = 1 \cdot (n-1) + 2 \cdot (n-2) + 3 \cdot (n-3) + \dots + (n-1) \cdot 1$. We can observe that the general term is of the form $r \cdot (n-r)$, where $r$ ranges from $1$ to $n-1$. So, $S_n = \sum_{r=1}^{n-1} r(n-r)$. Expanding the term inside the summation, we get $r(n-r) = nr - r^2$. Thus, $S_n = \sum_{r=1}^{n-1} (nr - r^2)$. Using the linearity of summation, we can split this into two sums: $S_n = n \sum_{r=1}^{n-1} r - \sum_{r=1}^{n-1} r^2$. Now, we apply the standard summation formulas for the sum of the first $k$ integers and the sum of the first $k$ squares, with $k = n-1$: $S_n = n \left( \frac{(n-1)((n-1)+1)}{2} \right) - \left( \frac{(n-1)((n-1)+1)(2(n-1)+1)}{6} \right)$ $S_n = n \left( \frac{(n-1)n}{2} \right) - \left( \frac{(n-1)n(2n-1)}{6} \right)$ $S_n = \frac{n^2(n-1)}{2} - \frac{n(n-1)(2n-1)}{6}$. To simplify, we find a common denominator (6) and factor out common terms: $S_n = \frac{3n^2(n-1)}{6} - \frac{n(n-1)(2n-1)}{6}$ $S_n = \frac{n(n-1)}{6} [3n - (2n-1)]$ $S_n = \frac{n(n-1)}{6} [3n - 2n + 1]$ $S_n = \frac{n(n-1)(n+1)}{6}$. Why this step? Deriving a closed-form expression for $S_n$ simplifies the subsequent calculations and allows us to work with a more manageable algebraic form.

Step 2: Simplify the General Term of the Infinite Sum. The general term of the infinite sum is $\left( \frac{2S_n}{n!} - \frac{1}{(n-2)!} \right)$. Substitute the derived expression for $S_n$: $2S_n = 2 \cdot \frac{n(n-1)(n+1)}{6} = \frac{n(n-1)(n+1)}{3}$. Now, substitute this into the general term: $\frac{n(n-1)(n+1)}{3n!} - \frac{1}{(n-2)!}$. We use the factorial property $n! = n(n-1)(n-2)!$ to simplify the first term: $\frac{n(n-1)(n+1)}{3 \cdot n(n-1)(n-2)!} - \frac{1}{(n-2)!}$. Cancel the common factors $n(n-1)$ in the first term: $\frac{n+1}{3(n-2)!} - \frac{1}{(n-2)!}$. To combine these terms, we ensure a common denominator, which is $3(n-2)!$: $\frac{n+1}{3(n-2)!} - \frac{3}{3(n-2)!}$. Combine the numerators: $\frac{(n+1) - 3}{3(n-2)!} = \frac{n-2}{3(n-2)!}$. Now, use the factorial property $(n-2)! = (n-2)(n-3)!$ (since $n \ge 4$, $n-2 \ge 2$, so this is valid): $\frac{n-2}{3(n-2)(n-3)!}$. Cancel the common factor $(n-2)$: $\frac{1}{3(n-3)!}$. Why this step? Simplifying the general term of the infinite sum into a compact form involving factorials is crucial for recognizing its connection to known series expansions.

Step 3: Evaluate the Infinite Sum. The infinite sum is $\sum_{n=4}^\infty \left( \frac{2S_n}{n!} - \frac{1}{(n-2)!} \right)$. Using the simplified general term from Step 2, the sum becomes: $\sum_{n=4}^\infty \frac{1}{3(n-3)!}$. Factor out the constant $\frac{1}{3}$: $\frac{1}{3} \sum_{n=4}^\infty \frac{1}{(n-3)!}$. To relate this to the Taylor series for $e$, we change the index of summation. Let $k = n-3$. When $n=4$, $k = 4-3 = 1$. As $n \to \infty$, $k \to \infty$. The sum transforms to: $\frac{1}{3} \sum_{k=1}^\infty \frac{1}{k!}$. The Taylor series for $e$ is $e = \sum_{k=0}^\infty \frac{1}{k!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \dots$. Our sum is $\sum_{k=1}^\infty \frac{1}{k!} = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$. This is the series for $e$ minus its $k=0$ term: $\sum_{k=1}^\infty \frac{1}{k!} = \left( \sum_{k=0}^\infty \frac{1}{k!} \right) - \frac{1}{0!}$. Since $\frac{1}{0!} = 1$, we have: $\sum_{k=1}^\infty \frac{1}{k!} = e - 1$. Substituting this back into our expression: $\frac{1}{3} (e-1) = \frac{e-1}{3}$. Why this step? By manipulating the summation index and recognizing the form of the Taylor series for $e$, we can evaluate the infinite sum.

Common Mistakes & Tips

• Index Errors: Be extremely careful when changing summation indices. Ensure the starting, ending, and incremental steps of the new index correctly map to the original index.
• Taylor Series for $e$: Remember that $e = \sum_{k=0}^\infty \frac{1}{k!} = 1 + \sum_{k=1}^\infty \frac{1}{k!}$. A common mistake is to equate $\sum_{k=1}^\infty \frac{1}{k!}$ directly with $e$ without subtracting the $k=0$ term.
• Algebraic Simplification: Errors in simplifying $S_n$ or the general term of the infinite sum will propagate. Double-check all algebraic manipulations, especially with factorials.

Summary The problem required finding the value of an infinite sum. We first derived a closed-form expression for $S_n$ using summation formulas. Then, we simplified the general term of the infinite sum by substituting the expression for $S_n$ and utilizing factorial properties. Finally, by recognizing the resulting series as a part of the Taylor expansion of $e$, we evaluated the infinite sum.

The final answer is $\boxed{\frac{e - 1}{3}}$ which corresponds to option (A).