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Radius and Interval of Convergence Calculator

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Example
Created on 2024-06-20Asked by Jacob Robinson (Solvelet student)
Determine the radius and interval of convergence for the series ∑n=1∞xnn \sum_{n=1}^{\infty} \frac{x^n}{n} .

Solution

To determine the radius and interval of convergence for the series ∑n=1∞xnn \sum_{n=1}^{\infty} \frac{x^n}{n} : 1. Use the Ratio Test to find the radius of convergence: lim⁡n→∞∣an+1an∣=lim⁡n→∞∣xn+1n+1xnn∣=lim⁡n→∞∣xn+1⋅nxn⋅(n+1)∣=lim⁡n→∞∣x⋅nn+1∣=∣x∣. \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{x^{n+1}}{n+1}}{\frac{x^n}{n}} \right| = \lim_{n \to \infty} \left| \frac{x^{n+1} \cdot n}{x^n \cdot (n+1)} \right| = \lim_{n \to \infty} \left| \frac{x \cdot n}{n+1} \right| = |x|. 2. Set the limit less than 1 to find the radius of convergence: ∣x∣<1. |x| < 1. Therefore, the radius of convergence is R=1 R = 1 . 3. To find the interval of convergence, test the endpoints x=±1 x = \pm 1 : - At x=1 x = 1 : ∑n=1∞1n(diverges). \sum_{n=1}^{\infty} \frac{1}{n} \quad \text{(diverges)}. - At x=−1 x = -1 : ∑n=1∞(−1)nn(converges by the Alternating Series Test). \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \quad \text{(converges by the Alternating Series Test)}. Therefore, the interval of convergence is −1≤x<1 -1 \leq x < 1 . Solved on Solvelet with Basic AI Model
Some of the related questions asked by Evelyn Thompson on Solvelet
1. Find the radius and interval of convergence for the power series ∑n=0∞(x−1)nn \sum_{n=0}^\infty \frac{(x - 1)^n}{n} .2. Determine the radius and interval of convergence for the power series ∑n=0∞(x+3)n2n+1\sum_{n=0}^{\infty} \frac{(x + 3)^n}{2n + 1}.
DefinitionThe radius of convergence of a power series is the distance over which the series converges to a limit. The interval of convergence is all the x-values for which the series will converge at. For instance, in the series ∑n=0∞n! Then from the geometric series analysis on xnxn​ the radius of convergence is ∞ that is we have convergence for all real x in −∞,∞−∞,∞.
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