- 56 名前：１３２人目の素数さん mailto:sage [2007/06/07(木) 01:28:47 ]
- >>15 [C.854] 別法
S_n = (n!)Σ[k=1,∞) (H_k)/[k(k+1)……(k+n)] とおき、
S_1 = (π^2)/6 と S_n - S_(n+1) = 1/(n^2) から
S_n = (π^2)/6 -1 -1/(2^2) - … - 1/{(n-1)^2}
を示す。
S_1 = Σ[k=1,∞) (H_k)/[k(k+1)]
= Σ[k=1,∞) (H_k){1/k - 1/(k+1)}
= Σ[k=1,∞) {(H_k)/k - H_(k-1)/k} (← H_0 = 0 )
= Σ[k=1,∞) 1/(k^2)
= ζ(2)
= (π^2)/6.
S_n - S_(n+1) = Σ[k=1,∞) (H_k){ (n!)/[k(k+1)…(k+n)] - (n+1)!/[k(k+1)…(k+n+1)] }
= (n!)Σ[k=1,∞) (H_k)/[(k+1)……(k+n+1)]
= (n-1)!Σ[k=1,∞) (H_k){1/[(k+1)…(k+n)] - 1/[(k+2)…(k+n+1)] }
= (n-1)!Σ[k=1,∞) { (H_k)/[(k+1)…(k+n)] - H_(k-1)/[(k+1)…(k+n)] } (← H_0 =0)
= (n-1)!Σ[k=1,∞) 1/[k(k+1)…(k+n)]
= (n-1)!Σ[k=1,∞) (1/n){ 1/[k(k+1)…(k+n-1)] - 1/[(k+1)…(k+n)] }
= (n-1)!(1/n)(1/n!)
= 1/(n^2).
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