Question

How do you test for convergence for `(sin 4n)/(4^n)` for n=1 to infinity?

Answer 1

Since the function sinus is with range `[-1,1]`, than:

`sin4n<=1` and so:

`sin(4n)/4^n<=1/4^n<=1/n^2` (for `n` big enough) that is a convergent series.

So our series is convergent for the principle of comparision.