Question:medium

What is the radius of convergence of the power series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \)?

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\(\sum \frac{x^n}{n!} = e^x\)
Exponential series converges for all real numbers \(x\)
Updated On: Mar 19, 2026
  • \(0\)
  • \(1\)
  • \(\infty\)
  • \(e\)
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The Correct Option is C

Solution and Explanation

This question asks for the radius of convergence of a well-known power series. The radius of convergence defines the interval on which the series converges to a function.
Step 1: Understanding the Question:
We need to find the value \(R\) such that the power series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) converges for all \(x\) with \(|x|<R\) and diverges for \(|x|>R\).
Step 2: Key Formula or Approach:
The most common method to find the radius of convergence is the Ratio Test. For a series \( \sum a_n x^n \), we compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). The radius of convergence is then \( R = \frac{1}{L} \).
Step 3: Detailed Explanation:

Identify the coefficient \(a_n\): In the given series, the coefficient of \(x^n\) is \( a_n = \frac{1}{n!} \).

Set up the ratio: We need to compute the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \). \[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{1/(n+1)!}{1/n!} = \frac{n!}{(n+1)!} = \frac{n!}{(n+1) \cdot n!} = \frac{1}{n+1} \]

Compute the limit L: \[ L = \lim_{n \to \infty} \frac{1}{n+1} = 0 \]

Calculate the radius of convergence R: \[ R = \frac{1}{L} = \frac{1}{0} \]

A limit of 0 implies that the radius of convergence is infinite. This means the series converges for all real (or complex) values of \(x\).
Step 4: Final Answer:
The radius of convergence is \( \infty \).
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