use the comparison test or limit comparison test to determine if the series converges.\n sum_{n =…

Explanation:

Step1: Choose a comparison series

For large (n), the dominant terms in the numerator and denominator of (\frac{n^{2}+4}{n^{7}+6}) are (n^{2}) and (n^{7}) respectively. So, we compare the given series (\sum_{n = 1}^{\infty}\frac{n^{2}+4}{n^{7}+6}) with the series (\sum_{n=1}^{\infty}\frac{n^{2}}{n^{7}}=\sum_{n = 1}^{\infty}\frac{1}{n^{5}}).

Step2: Apply the Limit - Comparison Test

We find the limit (\lim_{n\rightarrow\infty}\frac{\frac{n^{2}+4}{n^{7}+6}}{\frac{1}{n^{5}}}). [ \begin{align*} \lim_{n\rightarrow\infty}\frac{\frac{n^{2}+4}{n^{7}+6}}{\frac{1}{n^{5}}}&=\lim_{n\rightarrow\infty}\frac{(n^{2}+4)n^{5}}{n^{7}+6}\ &=\lim_{n\rightarrow\infty}\frac{n^{7}+4n^{5}}{n^{7}+6}\ &=\lim_{n\rightarrow\infty}\frac{1 + \frac{4}{n^{2}}}{1+\frac{6}{n^{7}}}\ &= 1 \end{align*} ] Since (0<1<\infty) and the series (\sum_{n = 1}^{\infty}\frac{1}{n^{5}}) is a (p -)series with (p = 5>1) (and (p -)series (\sum_{n=1}^{\infty}\frac{1}{n^{p}}) converges for (p>1)), by the Limit - Comparison Test, the series (\sum_{n = 1}^{\infty}\frac{n^{2}+4}{n^{7}+6}) converges.

Answer:

Converges