I think that your reasoning is correct because
n=1∑∞4n+cos2(n)2nsin2(5n)≤n=1∑∞4n2n
And
n=1∑∞4n2n is a convergent geometric series.
If you want to calculate the limit of the series do this:
n→∞lim4n+cos2(n)2nsin2(5n)=n→∞limenln(2/4)=n→∞limenln(1/2)=n→∞limen[ ln(1)−ln(2) ]=n→∞limen[−ln(2) ]=e−∞=0
We ignored
sin2(5n) and
cos2(n) because they are just a number between
0 and
1.