Question

question 11 of 13 > write an inequality comparing $\frac{n^{2}-6}{n^{5}+6}$ with $\frac{1}{n^{3}}$ for $ngeq1$. (express numbers in exact form. use symbolic notation and fractions where needed.) inequality: use this inequality to draw a conclusion about the series $sum_{n = 1}^{infty}\frac{n^{2}-6}{n^{5}+6}$. the series diverges by the direct comparison test because $sum_{n = 1}^{infty}\frac{1}{n^{3}}$ also diverges. it is not possible to draw a conclusion about the convergence of the series because $sum_{n = 1}^{infty}\frac{1}{n^{3}}$ diverges. the series converges by the direct comparison test because $sum_{n = 1}^{infty}\frac{1}{n^{3}}$ also converges.

Explanation:

Step1: Cross - multiply to compare

For \(n\geq1\), we want to compare \(\frac{n^{2}-6}{n^{5}+6}\) and \(\frac{1}{n^{3}}\). Cross - multiply: \((n^{2}-6)n^{3}\) and \(n^{5}+6\).
\((n^{2}-6)n^{3}=n^{5}-6n^{3}\).

Step2: Analyze the difference

Find the difference between \(n^{5}+6\) and \(n^{5}-6n^{3}\): \((n^{5}+6)-(n^{5}-6n^{3}) = 6 + 6n^{3}>0\) for \(n\geq1\). So \(n^{5}+6>(n^{2}-6)n^{3}\). Then \(\frac{n^{2}-6}{n^{5}+6}<\frac{1}{n^{3}}\) for \(n\geq1\).

Step3: Use the Direct Comparison Test

The \(p\) - series \(\sum_{n = 1}^{\infty}\frac{1}{n^{p}}\) diverges when \(p = 3>1\). Since \(0<\frac{n^{2}-6}{n^{5}+6}<\frac{1}{n^{3}}\) for \(n\geq1\) and \(\sum_{n=1}^{\infty}\frac{1}{n^{3}}\) diverges, we cannot draw a conclusion about the convergence of \(\sum_{n = 1}^{\infty}\frac{n^{2}-6}{n^{5}+6}\) using the Direct Comparison Test in this case.

Answer:

inequality: \(\frac{n^{2}-6}{n^{5}+6}<\frac{1}{n^{3}}\)
It is not possible to draw a conclusion about the convergence of the series because \(\sum_{n = 1}^{\infty}\frac{1}{n^{3}}\) diverges.