<p>The value of \(\displaystyle\sum_{n=0}^{\infty} \dfrac{(\ln x)^n}{n!}\) is:</p>
Step-by-Step Solution
Key Concept: Recognize that the given sum matches the Taylor series expansion of e^u where u = ln(x). The sum ∑(u^n/n!) from n=0 to ∞ is the definition of e^u.
<p><strong>Step 1:</strong> Recognize the standard form. The sum ∑(u^n/n!) from n=0 to ∞ is the Taylor series expansion of e^u.</p><p><strong>Step 2:</strong> Identify u = ln(x) in our series. We have:</p><p>∑_{n=0}^{∞} (ln x)^n/n! = e^(ln x)</p><p><strong>Step 3:</strong> Apply the fundamental logarithmic identity: e^(ln x) = x (for x > 0).</p><p><strong>Step 4:</strong> Therefore, the sum equals <strong>x</strong>.</p><p>∴ Answer: <strong>C (which is x)</strong></p>
Correct Answer: C