Comparison and Limit Comparison Tests — Question 5

Section 11-7 · Calculus pages 2069–2070

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*Question 5 Consider ∑n=1∞n3n5+4\displaystyle\sum_{n=1}^{\infty}\frac{n^3}{n^5+4}.

  1. Choose a benchmark from the dominant powers.

  2. Compute the limiting ratio and verify all Limit Comparison Test conditions.

  3. Classify the series and give a direct upper bound.

*Question 5 – Solution

The leading quotient n3/n5=1/n2n^3/n^5=1/n^2 suggests bn=1/n2b_n=1/n^2, a convergent pp-series.

limn→∞n3/(n5+4)1/n2=limn→∞n5n5+4=1.\lim_{n\to\infty}\frac{n^3/(n^5+4)}{1/n^2} =\lim_{n\to\infty}\frac{n^5}{n^5+4}=1. All terms are positive and the limit lies in (0,∞)(0,\infty), so Limit Comparison proves convergence.

Since n5+4≥n5n^5+4\ge n^5, 0<n3n5+4≤1n2.0<\frac{n^3}{n^5+4}\le\frac1{n^2}. This upper comparison with a convergent series gives the same conclusion.