And bn =1/2 n.

| an | = | sin n|/2^n<； = 1/2^n=bn

The series bn converges because it is a geometric series with equal ratio 1/2.

Therefore, through comparison and judgment, the series an is absolutely convergent.

The second part:

because

Sin k = im (e (ik))

Sam sink/2 k

= sum of im (e (ik))/2 k

Commutative sum and imaginary part operation

=Im [sum e^(ik)/2^k]

=Im [sum (e^i/2)^k)

Because | E I/2 | = | E I |/2 = 1/2 is a convergent geometric series, the equation ratio is E I/2, and the first term is 1.

= the first item /( 1- common ratio)

= 1/( 1-e^i/2)

Note that e I = cos 1+isin 1.

= 2/[2-(cos 1+isin 1)]

= 2/[(2-cos 1)-isin 1]

=2[(2-cos 1)+isin 1]/[(2-cos 1)^2+(-sin 1)^2]

= 2[(2-cos 1)+is in 1]/[5-4 cos 1]

Take the imaginary part and get it.

2sin 1/(5-4cos 1)