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)