Answer:
It means
also converges.
Explanation:
The actual Series is::

The method we are going to use is comparison method:
According to comparison method, we have:

If series one converges, the second converges and if second diverges series, one diverges
Now Simplify the given series:
Taking"n^2"common from numerator and "n^6"from denominator.
![=(n^2[7-(4)/(n)+(3)/(n^2)])/(n^6[(12)/(n^6)+2]) \\\\=([7-(4)/(n)+(3)/(n^2)])/(n^4[(12)/(n^6)+2])](https://img.qammunity.org/2021/formulas/mathematics/college/fhc5pz8w7vp78fe6khgrnztyvz5ce0whop.png)
![\sum_(n=1)^(inf)a_n=\sum_(n=1)^(inf)([7-(4)/(n)+(3)/(n^2)])/([(12)/(n^6)+2])\ \ \ \ \ \ \ \ \sum_(n=1)^(inf)b_n=\sum_(n=1)^(inf) (1)/(n^4)](https://img.qammunity.org/2021/formulas/mathematics/college/juqnatcgxlrt41ugnul3u229zvz2bxxwii.png)
Now:
![\sum_(n=1)^(inf)a_n=\sum_(n=1)^(inf)([7-(4)/(n)+(3)/(n^2)])/([(12)/(n^6)+2])\\ \\\lim_(n \to \infty) a_n = \lim_(n \to \infty) ([7-(4)/(n)+(3)/(n^2)])/([(12)/(n^6)+2])\\=(7-(4)/(inf)+(3)/(inf))/((12)/(inf)+2)\\\\=(7)/(2)](https://img.qammunity.org/2021/formulas/mathematics/college/415vlygfvt9i9pqr9b74iawm6b62a8wolc.png)
So a_n is finite, so it converges.
Similarly b_n converges according to p-test.
P-test:
General form:

if p>1 then series converges. In oue case we have:

p=4 >1, so b_n also converges.
According to comparison test if both series converges, the final series also converges.
It means
also converges.