∑(n = 24 to ∞) n² / (n³ + 6)^(3/2) (a) assume the conditions for the integral test are met. use the integral…

Explanation:

Step1: Identify the function for integral test

Let (f(x)=\frac{x^{2}}{(x^{3}+6)^{\frac{3}{2}}}), then the associated improper - integral for the series (\sum_{n = 24}^{\infty}\frac{n^{2}}{(n^{3}+6)^{\frac{3}{2}}}) is (\int_{24}^{\infty}\frac{x^{2}}{(x^{3}+6)^{\frac{3}{2}}}dx).

Step2: Use substitution

Let (u=x^{3}+6), then (du = 3x^{2}dx) and (x^{2}dx=\frac{1}{3}du). When (x = 24), (u=24^{3}+6), and as (x\rightarrow\infty), (u\rightarrow\infty). So the integral becomes (\frac{1}{3}\int_{24^{3}+6}^{\infty}u^{-\frac{3}{2}}du).

Step3: Evaluate the integral

Using the power - rule for integration (\int u^{r}du=\frac{u^{r + 1}}{r+1}+C) ((r\neq - 1)), for (r=-\frac{3}{2}), we have (\frac{1}{3}\int_{24^{3}+6}^{\infty}u^{-\frac{3}{2}}du=\frac{1}{3}\lim_{b\rightarrow\infty}\int_{24^{3}+6}^{b}u^{-\frac{3}{2}}du=\frac{1}{3}\lim_{b\rightarrow\infty}\left[- 2u^{-\frac{1}{2}}\right]{24^{3}+6}^{b}). [ \begin{align*} &=\frac{1}{3}\lim{b\rightarrow\infty}\left(-\frac{2}{\sqrt{b}}+\frac{2}{\sqrt{24^{3}+6}}\right)\ &=\frac{2}{3\sqrt{24^{3}+6}} \end{align*} ] Since the improper integral (\int_{24}^{\infty}\frac{x^{2}}{(x^{3}+6)^{\frac{3}{2}}}dx) converges, by the Integral Test, the series (\sum_{n = 24}^{\infty}\frac{n^{2}}{(n^{3}+6)^{\frac{3}{2}}}) converges.

Step4: Determine other appropriate tests

The Limit Comparison Test can also be used. We can compare the series (\sum_{n = 24}^{\infty}\frac{n^{2}}{(n^{3}+6)^{\frac{3}{2}}}) with the series (\sum_{n = 24}^{\infty}\frac{n^{2}}{(n^{3})^{\frac{3}{2}}}=\sum_{n = 24}^{\infty}\frac{n^{2}}{n^{\frac{9}{2}}}=\sum_{n = 24}^{\infty}n^{2-\frac{9}{2}}=\sum_{n = 24}^{\infty}n^{-\frac{5}{2}}), which is a (p) - series with (p=\frac{5}{2}>1).

Answer:

(a) (\int_{24}^{\infty}\frac{x^{2}}{(x^{3}+6)^{\frac{3}{2}}}dx=\frac{2}{3\sqrt{24^{3}+6}}), converges (b) Limit Comparison Test