Determine the convergence or divergence of the series.

sigma 1/n^2 (n+3)
n=1

where n goes from 1 to infinity.

its sigma 1/[(n^2)(n+3)]

Is it 1/[(n^2)(n+3)] or is it (n+3)/(n^2) ?
If it is: (n+3)/(n^2) then S = ∑1/n + 3∑1/n^2
As ∑1/n diverges(harmonic sum), S = ∑1/n + 3∑1/n^2 diverges

If it is 1/[(n^2)(n+3)]:
(n+3)^2 - n^2 = 6n + 9
.'. 1 = (n+3)^2 - n^2 - 6n - 8
.'. ∑1/[(n^2)(n+3)] = ∑(n+3)/n^2 - ∑1/(n+3) - 6∑1/[n(n+3)] - 8∑1/[(n^2)(n+3)]
or, 9∑1/[(n^2)(n+3)] = ∑1/n + 3∑1/n^2 - ∑1/(n+3) - 6∑1/[n(n+3)]

I now take up: 6∑1/[n(n+3)]
= 6/3 ∑[(n+3)-n]/[n(n+3)]
= 2 ∑1/n - 2 ∑1/(n+3)

.'. 9∑1/[(n^2)(n+3)] = ∑1/n + 3∑1/n^2 - ∑1/(n+3) - 2 ∑1/n + 2 ∑1/(n+3)
or, 9∑1/[(n^2)(n+3)] = ∑1/(n+3) - ∑1/n + 3∑1/n^2
Now, ∑1/(n+3) - ∑1/n is a small -ve quantity
It is not going to affect your answer and your sum will only depend on ∑1/n^2

.'. 9 ∑1/[(n^2)(n+3)] < 3∑1/n^2
or, ∑1/[(n^2)(n+3)] < (1/3)∑1/n^2

your series, if the second one will depend on what you think of ∑1/n^2 series!