Question

求极限，如图

Answer 1

limn→∞(sinπ/n+sin2π/n+...+sinπ)
=Ssinxdx/(π/n)丨(0~π)
=-ncosx/π丨(0~π)
=2n/π，
limn→∞[(sinπ/n)/n+(sin2π/n)/n+...+(sinπ)/n]
=limn→∞(sinπ/n+sin2π/n+...+sinπ)/n
=limn→∞(2n/π)/n
=2/π，
limn→∞[(sinπ/n)/(n+1)+(sin2π/n)/(n+1)+...+(sinπ)/(n+1)]
=limn→∞(sinπ/n+sin2π/n+...+sinπ)/(n+1)
=limn→∞(2n/π)/(n+1)
=2/π，
limn→∞[(sinπ/n)/n+(sin2π/n)/n+...+(sinπ)/n]
>limn→∞[(sinπ/n)/(n+1)+(sin2π/n)/(n+1/2)+...+(sinπ)/(n+1/n)]
>limn→∞[(sinπ/n)/(n+1)+(sin2π/n)/(n+1)+...+(sinπ)/(n+1)]
由夹逼定理：
limn→∞[(sinπ/n)/(n+1)+(sin2π/n)/(n+1/2)+...+(sinπ)/(n+1/n)]=2/π。