The three series ${\displaystyle \underset{?}{\overset{?}{}}}{A}_n,{\displaystyle \underset{?}{\overset{?}{}}}{B}_n,$ and ${\displaystyle \underset{?}{\overset{?}{}}}{C}_n$ have terms

$A_n=\frac{1}{n^6},B_n=\frac{1}{n^3},C_n=\frac{1}{n}.$

Use the Limit Comparison Test to compare the following series to any of the above series. For each of the series below, you must enter two letters. The first is the letter $(A,B,$ or $C)$ of the series above that it can be legally compared to with the Limit Comparison Test. The second is $C$ if the given series converges, or $D$ if it diverges. So for instance, if you believe the series converges and can be compared with series C above, you would enter CC; or if you believe it diverges and can be compared with series $A,$ you would enter AD$.$

${\displaystyle \underset{n=1}{\overset{}{}}}\frac{3n^3+n^6}{748n^9+8n^3+6}$

$2{\displaystyle \underset{n=1}{\overset{}{}}}\frac{3n^3+n^2-3n}{8n^9-6n^8+4}$

${\displaystyle \underset{n=1}{\overset{}{}}}\frac{6n^2+3n^5}{3n^6+8n^3-3}$