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Root Test Calculator

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Example
Created on 2024-06-20Asked by Victoria Anderson (Solvelet student)
Determine the convergence of the series ∑n=1∞(n2n+1)n\sum_{n=1}^{\infty} \left(\frac{n}{2n+1}\right)^n using the root test.

Solution

To determine the convergence of the series ∑n=1∞(n2n+1)n\sum_{n=1}^{\infty} \left(\frac{n}{2n+1}\right)^n using the root test: 1. **Apply the root test:** lim⁡n→∞annwherean=(n2n+1)n. \lim_{n \to \infty} \sqrt[n]{a_n} \quad \text{where} \quad a_n = \left(\frac{n}{2n+1}\right)^n. 2. **Calculate the limit:** ann=(n2n+1). \sqrt[n]{a_n} = \left(\frac{n}{2n+1}\right). lim⁡n→∞(n2n+1)=lim⁡n→∞12+1n=12. \lim_{n \to \infty} \left(\frac{n}{2n+1}\right) = \lim_{n \to \infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2}. 3. **Conclusion:** Since lim⁡n→∞ann=12<1\lim_{n \to \infty} \sqrt[n]{a_n} = \frac{1}{2} < 1, the series ∑n=1∞(n2n+1)n\sum_{n=1}^{\infty} \left(\frac{n}{2n+1}\right)^n converges by the root test. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Mila Garcia on Solvelet
1. Determine the convergence or divergence of the series ∑n=1∞n!nn \sum_{n=1}^\infty \frac{n!}{n^n} .2. Find the radius of convergence for the power series ∑n=0∞xnn! \sum_{n=0}^{\infty} \frac{x^n}{n!} .,
DefinitionThe Root Test- For testing the convergence of the infinite series.
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