<p>Find the sum: \[1 + \dfrac{1 + \dfrac{1}{1!}}{2} + \dfrac{1 + \dfrac{1}{1!} + \dfrac{1}{2!}}{2^2} + \dfrac{1 + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!}}{2^3} + \ldots \text{ to } \infty.\]</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator of the nth term approaches e as n→∞, and express each term using partial sums of the e series: the nth term is (1/2^n)·∑(k=0 to n-1) 1/k!. Interchange summation order to evaluate the double sum.
<p><strong>Step 1:</strong> Rewrite the series. The numerator of the nth term (n≥1) is the partial sum S_n = ∑(k=0 to n-1) 1/k!. So:</p><p>∑(n=1 to ∞) S_n/2^n = ∑(n=1 to ∞) [∑(k=0 to n-1) 1/k!]/2^n</p><p><strong>Step 2:</strong> Convert to double summation with correct limits. Writing out the terms:</p><p>= (1/1!)/2 + (1/0! + 1/1!)/2² + (1/0! + 1/1! + 1/2!)/2³ + ...</p><p>In double sum form: ∑(n=1 to ∞) ∑(k=0 to n-1) (1/k!)·(1/2^n)</p><p><strong>Step 3:</strong> Reindex by fixing k and letting n vary. For fixed k, we need n > k (so that k ≤ n-1). Thus:</p><p>∑(k=0 to ∞) ∑(n=k+1 to ∞) (1/k!)·(1/2^n) = ∑(k=0 to ∞) (1/k!) · ∑(n=k+1 to ∞) 1/2^n</p><p><strong>Step 4:</strong> Evaluate the inner geometric series:</p><p>∑(n=k+1 to ∞) 1/2^n = 1/2^(k+1) + 1/2^(k+2) + ... = (1/2^(k+1))/(1 - 1/2) = 1/2^k</p><p><strong>Step 5:</strong> Substitute back:</p><p>∑(k=0 to ∞) (1/k!) · (1/2^k) = ∑(k=0 to ∞) 1/(k!·2^k) = ∑(k=0 to ∞) (1/2)^k/k! = e^(1/2)</p><p><strong>Step 6:</strong> Add the initial term (1) from the original series:</p><p>∴ Answer: 1 + e^(1/2) = <strong>1 + √e</strong></p>
Correct Answer: 1