>>427
それ、不等号が逆

>>425
sin x > 2x/π (0<x<π/2)
を用いて
(n+1)π∫[0,π] < 2(n+1)π∫[0,π/2]
< 2(n+1)π∫[0,π/2] 1/(1+(nπ)^6 (2x/π)^2) dx
< 2(n+1)π∫[0,∞] 1/(1+(nπ)^6 (2x/π)^2) dx
< (n+1)/(2n^3)
≦ 1/n^2