Let fp(α)=eiαp2.e2iαp2.e3iαp2.e4iαp2…eiαp, (where i=√−1 and p∈N) then limn→∞fn(π) is
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fp(α)=eiαp2(1+2+3+4+…+p)=eiαp2.p(p+1)2=eiα2(1+1p)∴ limn→∞fn(π)=limn→∞eiπ2(1+1n)=eiπ2(1+0)=eiπ2=cosπ2+i sinπ2=i