find all cluster points of A sub n n1 inf A sub n12n 1ncos n

find all cluster points of {A sub n}
n=1-> inf
A sub n=(1/(2^n))+ ((-1)^n)cos ((n(pi))/4)+sin((n (pi))/2)

Solution

The cluster points will be obtained for n=4k, 4k+1, 4k+2, 4k+3.

For n=4k, A_n= 1/2^(4k)+cos(4k/4)+sin(4k/2)

= 1/2^(4k)+cos(k)+sin(2k)

Now, lim(n-> infinity) A_n =1 of r k=even (-1) for k=odd

For n=4k+1,

A_(4k+1) = 1/2^(4k+1) + (-1)^(4k+1) cos(4k+1)/4 + sin(4k+1)/2

=1/2^(4k+1) - cos(k+/4) + sin(2k+/2)

--> 1/2 +1 for k=even and -1/2 +1 for k=odd

For n=4k+2,

A_(4k+2) = 1/2^(4k+2) + (-1)^(4k+2) cos(4k+2)/4 + sin(4k+2)/2

=1/2^(4k+2) + cos(2k+1)/2 + sin(2k+1)

--> 0

For n=4k+3,

A_(4k+3) = 1/2^(4k+3) + (-1)^(4k+3) cos(4k+3)/4 + sin(4k+3)/2

= 1/2^(4k+3) - cos(k+3/4) + sin(2k+3/2)

--> (-1)+1/2 for k=even and (-1)-1/2 for k=odd

So all the Cluster points of A_n are:

0, 1, (-1), ±1±1/2