>>915 >>917
〔問題〕
次の不等式が成り立つことを示せ。
(1)　0 < x < π/2 とするとき、(2/π)x < sin(x) < x,
(2)　πlog(2) < π/2 + ∫[0,π/2] log(1+sin(x))dx < (1+π/2)log(1+π/2),

πlog(2) = 2.1775860903････
π/2 + ∫[0,π/2] log(1+sin(x))dx = 2γ + (π/2){1-log(2)} = 2.3139344670････
(1+π/2)log(1+π/2) = 2.4273862679････
ただし γ = 0.5772156649････

>>918
(1)　n=1