Root Test — Question 6

Section 11-11 · Calculus pages 2151–2152

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*Question 6 Analyze ∑n=1∞(n+12n)n\displaystyle\sum_{n=1}^{\infty}\left(\frac{n+1}{2n}\right)^n using the Root Test. State the root limit and the convergence classification.

*Question 6 – Solution

|(n+12n)n|n=n+12n=12+12n.\sqrt[n]{\left|\left(\frac{n+1}{2n}\right)^n\right|}=\frac{n+1}{2n}=\frac12+\frac1{2n}.

L=limn→∞(12+12n)=12<1.L=\lim_{n\to\infty}\left(\frac12+\frac1{2n}\right)=\frac12<1.

The series converges absolutely by the Root Test. Although the base varies with nn, it approaches 1/21/2, so the terms eventually have geometric-scale decay. For example, once n≥5n\ge5, the root is at most 3/53/5, so an≤(3/5)na_n\le(3/5)^n.