Comparison and Limit Comparison Tests — Question 5

Question 5 – Solution

Step 1: Select the benchmark.

The leading quotient $n^3/n^5=1/n^2$ suggests $b_n=1/n^2$, a convergent $p$-series.

Step 2: Compute the ratio limit.

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

Step 3: Verify directly.

Since $n^5+4\ge n^5$, $$0<\frac{n^3}{n^5+4}\le\frac1{n^2}.$$ This upper comparison with a convergent series gives the same conclusion.