When to use direct and limit comparison test?

if the limit comparison test yields an answer then so will a direct comparison test...

just might need a bit of algebra and logic

Ex : Σ { n = 1 , 2, ..} 1 / [ n² - 4n + 5 ] converges by lim test with Σ1 / n²......

but 1 / [ n² - 4 n + 5 ] > 1 / n² for n = 1,2,3,.. so direct won't work with that series �� 1 / n²...

however Convergence / Divergence is a " tail property "....

so let us use Σ { n = 3,4,... } 1 / [ n - 2]² ≡ Σ { n=1,2,3..} 1 / n² vs Σ {n=3,4,..} 1 / [ n² - 4 n + 5].....

now 1 / [ n² - 4n + 5 ] < 1 / [ n - 2 ] ² so since Σ 1/n² con verges then so does Σ 1 / [ n - 2 ]²

and thus Σ 1 / [ n² - 4n + 5 ] converges using a ' direct ' comparison...

Ex : Σ sin² n / [ n² ] using Σ 1/ n²..here the lim test will not work but sin² n / [ n² ] ≤ 1/ n²

so the direct comp test implies the 'unknown ' series converges