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Example
Created on 2024-06-20Asked by Jackson Scott (Solvelet student)
Determine whether the sequence an=2nn!a_n = \frac{2^n}{n!} converges or diverges.

Solution

To determine whether the sequence an=2nn!a_n = \frac{2^n}{n!} converges or diverges: 1. **Consider the ratio of successive terms:** an+1an=2n+1/(n+1)!2n/n!=2⋅2n/(n+1)!2n/n!=2n+1. \frac{a_{n+1}}{a_n} = \frac{2^{n+1} / (n+1)!}{2^n / n!} = \frac{2 \cdot 2^n / (n+1)!}{2^n / n!} = \frac{2}{n+1}. 2. **Take the limit as n→∞n \to \infty:** lim⁡n→∞2n+1=0. \lim_{n \to \infty} \frac{2}{n+1} = 0. 3. **Conclusion:** Since lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, the sequence an=2nn!a_n = \frac{2^n}{n!} converges to 0. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Jackson Scott on Solvelet
1. Determine whether the series ∑n=1∞1n \sum_{n=1}^\infty \frac{1}{n} converges or diverges.2. Find the sum of the series ∑n=1∞2n3n−1 \sum_{n=1}^{\infty} \frac{2^n}{3^{n-1}} .,
DefinitionSequence: Series: An Addendum to the Discussion on Sequence and Series Example: The series or summation of the sequence 1,2,3,4,5,……. is 1+2+3+4+5+….
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