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Question

Let fp(β)=(cosβp2+isinβp2)(cos2βp2+isin2βp2)....(cosβp+isinβp)
then limnfn(π)=.....

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Solution

In the last term write βpaspβp2
fp(β)=eiβ/p2+2iβ/p2++piβ/p2
=e(iβ/p2){1+2+3++p}=e(iβ/p2)p(p+1)2
=eiβ21+1p
fn(π)=eiπ21+1n
limnfn(π)=eiπ/2=cosπ2+isinπ2=i

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