Root Test — Question 1

Section 11-11 · Calculus pages 2141–2142

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*Question 1 Consider ∑n=1∞(2n+1)n5nnn.\sum_{n=1}^{\infty}\frac{(2n+1)^n}{5^n n^n}.

  1. Rewrite the summand as a single quantity raised to the nnth power.

  2. Apply the Root Test and classify the series.

*Question 1 – Solution

Let an=(2n+1)n5nnn=(2n+15n)n=(25+15n)n.a_n=\frac{(2n+1)^n}{5^n n^n}=\left(\frac{2n+1}{5n}\right)^n=\left(\frac25+\frac1{5n}\right)^n.

Since an>0a_n>0, |an|n=2n+15n=25+15n.\sqrt[n]{|a_n|}=\frac{2n+1}{5n}=\frac25+\frac1{5n}. Therefore L=lim⁡n→∞|an|n=2/5<1\displaystyle L=\lim_{n\to\infty}\sqrt[n]{|a_n|}=2/5<1.

The series converges absolutely by the Root Test. The slowly varying 1/(5n)1/(5n) correction disappears in the limit, leaving geometric-scale decay with factor 2/52/5.