Question

证明级数收敛问题

Answer 1

a[n]+a[n+2] = ∫{0,π/4} (tan(x))^n dx+∫{0,π/4} (tan(x))^(n+2) dx
= ∫{0,π/4} (tan(x))^n·(1+tan²(x)) dx
= ∫{0,π/4} (tan(x))^n·(1/cos²(x)) dx
= ∫{0,π/4} (tan(x))^n·(tan(x))' dx
= (tan(π/4))^(n+1)/(n+1)-(tan(0))^(n+1)/(n+1)
= 1/(n+1).
因此(a[n]+a[n+2])/n = 1/(n(n+1)) = 1/n-1/(n+1).
∑{1 ≤ n} (a[n]+a[n+2])/n = ∑{1 ≤ n} (1/n-1/(n+1)) = 1.
由a[n] = ∫{0,π/4} (tan(x))^n dx > 0,∑a[n]/n^λ是正项级数.
又a[n] = 1/(n+1)-a[n+2] < 1/(n+1) < 1/n,故a[n]/n^λ < 1/n^(1+λ).
而当λ > 0,级数∑1/n^(1+λ)收敛,根据比较判别法,∑a[n]/n^λ也收敛.