Proving the convegence of the series sqrt(n)/(n^2 - 3)

For very large n, you can approximate
sqrt(n)/(n^2-3) ~= sqrt(n)/(n^2) = 1/n^3/2 = b_n
(as n approaches infinity, -3's effect approaches 0)

If you don't want to be like "for very large n", you can also observe that
sqrt(n)/(n^2-3) < sqrt(n)/(n^2/2)
for n>2 this holds, so you can squeeze the terms under 1/2*n^(-3/2) which is half a convergent p-series.
n=1 and n=2 are not a problem cause there are only a finite number of them, and none are infinite.


It'd be simpler on something like sqrt(n)/(n^2+2) because you can directly observe
sqrt(n)/(n^2+2) < sqrt(n)/(n^2)
for any n>0.