<p>\(\displaystyle\sum_{n=0}^{\infty} \dfrac{(\log_e x)^n}{n!}\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize this as the Taylor series expansion of e^u where u = log_e(x). The sum ∑(u^n/n!) from n=0 to ∞ equals e^u, so the answer is e^(log_e x) = x.
<p><strong>Step 1:</strong> Identify the series pattern. We have ∑(n=0 to ∞) [log_e(x)]^n / n!, which matches the standard form ∑(n=0 to ∞) u^n / n! where u = log_e(x).</p><p><strong>Step 2:</strong> Recall that the Taylor series for e^u is: e^u = ∑(n=0 to ∞) u^n / n!. This is a fundamental series that converges for all real u.</p><p><strong>Step 3:</strong> Apply the formula with u = log_e(x): ∑(n=0 to ∞) [log_e(x)]^n / n! = e^(log_e(x))</p><p><strong>Step 4:</strong> Simplify using the property that e^(log_e(x)) = x (the exponential and natural logarithm are inverse functions).</p><p>∴ Answer: <strong>x</strong></p>
Correct Answer: B