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Ratio Test Calculator

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Example
Created on 2024-06-20Asked by Jackson Williams (Solvelet student)
Use the Ratio Test to determine the convergence or divergence of the series ∑n=1∞2nn! \sum_{n=1}^{\infty} \frac{2^n}{n!} .

Solution

To determine the convergence or divergence of the series ∑n=1∞2nn! \sum_{n=1}^{\infty} \frac{2^n}{n!} using the Ratio Test: 1. **Calculate the ratio of successive terms:** lim⁡n→∞∣an+1an∣=lim⁡n→∞∣2n+1(n+1)!2nn!∣=lim⁡n→∞∣2n+1⋅n!2n⋅(n+1)!∣=lim⁡n→∞∣2⋅2n⋅n!2n⋅(n+1)⋅n!∣=lim⁡n→∞∣2n+1∣. \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} \right| = \lim_{n \to \infty} \left| \frac{2^{n+1} \cdot n!}{2^n \cdot (n+1)!} \right| = \lim_{n \to \infty} \left| \frac{2 \cdot 2^n \cdot n!}{2^n \cdot (n+1) \cdot n!} \right| = \lim_{n \to \infty} \left| \frac{2}{n+1} \right|. 2. **Simplify the expression:** lim⁡n→∞∣2n+1∣=0. \lim_{n \to \infty} \left| \frac{2}{n+1} \right| = 0. 3. **Interpret the result:** Since the limit is less than 1, the Ratio Test confirms that the series ∑n=1∞2nn! \sum_{n=1}^{\infty} \frac{2^n}{n!} converges. Therefore, the series converges. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Liam Anderson on Solvelet
1. Determine the convergence or divergence of the series ∑n=1∞n!2n! \sum_{n=1}^\infty \frac{n!}{2n!} .2. Find the radius of convergence for the power series ∑n=0∞xnn2 \sum_{n=0}^{\infty} \frac{x^n}{n^2} .
DefinitionRatio Test: Used to check the absolute convergence of a series. For a series ∑an, it states that if (\lim_{n \to \infty} \left
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