Question

Let \( f \) be a real-valued function of a real variable, such that \( |f^{(n)}(0)| \leq K \) for all \( n \in \mathbb{N} \), where \( K > 0. \) Which of the following is/are true?
 

• A. \( \dfrac{|f^{(n)}(0)|^{1/n}}{n!} \to 0 \) as \( n \to \infty \)
• B. \( \dfrac{|f^{(n)}(0)|^{1/n}}{n!} \to \infty \) as \( n \to \infty \)
• C. \( f^{(n)}(x) \) exists for all \( x \in \mathbb{R} \) and all \( n \in \mathbb{N} \)
• D. \( \sum_{n=1}^{\infty} \dfrac{f^{(n)}(0)}{(n-1)!} \) is absolutely convergent

Hint:

Factorials dominate exponential growth; terms involving \( 1/n! \) often guarantee convergence.

Answer 1

The Correct Option is A, D

Step 1: Analyze (A).
Since \( |f^{(n)}(0)| \le K \), \[ \dfrac{|f^{(n)}(0)|^{1/n}}{n!} \le \dfrac{K^{1/n}}{n!}. \] As \( n! \to \infty \) much faster than \( K^{1/n} \), the term tends to 0. Hence (A) is true.

Step 2: Analyze (D).
We have \[ \left| \frac{f^{(n)}(0)}{(n-1)!} \right| \le \frac{K}{(n-1)!}. \] Since \( \sum \frac{1}{(n-1)!} \) converges, by comparison test, the series converges absolutely. Thus, (D) is true.

Step 3: Analyze (B) and (C).
(B) contradicts (A), so false. (C) cannot be deduced from the given condition at a single point \( 0 \).

Final Answer: \[ \boxed{\text{(A) and (D)}} \]