1. arctan(n)<pi/2
So use comparison test with the conv pseries 1/n^3
2. rewrite (-3/4)^n - (1/4)^n the sum of 2 convergent geometric series
I'll have a think about these, but for 1., I would say you could use a comparison test (could someone verify?)
Since we can say that
, which is a p-series.
This p-series converges to a number, and since the original series is less than this number, the original series must also converge.