Question Statement:
The sum of the following infinite series is:
1
1! + 1
2! + 1
3! + 1
4! + 1
5! + ⋯
Options:
-
(A) 𝜋
-
(B) 1 + e
-
(C) e - 1
-
(D) e
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2025 Q.3)**
**1. Governing Concept**:
- **Topic**: Engineering Mathematics & Taylor Series of Exponential Function
- The Maclaurin series expansion of the exponential function $e^x$ is:
$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
- Substituting $x = 1$:
$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$
- The given series starts from $n = 1$:
$S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots = e - 1$
**2. Option Analysis**:
- **Option (A) [$e$] is INCORRECT**: Fails to subtract the initial term $1/0! = 1$.
- **Option (B) [$e + 1$] is INCORRECT**: Adds 1 instead of subtracting.
- **Option (C) [$e - 1$] is CORRECT**: Exactly equals $e - 1 \approx 2.71828 - 1 = 1.71828$.
- **Option (D) [$1/e$] is INCORRECT**: Represents $e^{-1}$.
**3. Final Answer**:
- **Correct Choice**: **C**
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