Succession limit

-1<=cos(inf)<=1, 2^inf=inf, cos(inf)/inf=0. Q.E.D.

As n -> $\infty$ cos(n) -> 1

Hence $\frac { 1 }{ { 2 }^{ n } } \Longrightarrow \frac {1}{\infty} = 0$

cos(n) will only return a value with absolute value 1 while 2^n will balloon up

Exhaust the cosine, They are so small numbers that are 0