>236
　1/{m^p + (m^q)x^2} ≦ 1/(m^p) ≦ ∫[m-1/2,m+1/2] 1/(x^p) dx = (1/(p-1)){1/[(m-1/2)^(p-1)] - 1/[(m+1/2)^(p-1)]},
　n>N　⇒　0 < f_n(x) - f_N(x) = Σ[m=N+1,n] 1/{m^p + (m^q)x^2} < (1/(p-1)){1/(N-1/2)^(p-1)},
n→∞ として
　0 ≦ f(x) - f_N(x) = Σ[m=N+1,∞) 1/{m^p + (m^q)x^2} ≦ (1/(p-1)){1/(N-1/2)^(p-1)},
任意のε>0 に対して N = [ {1/(p-1)ε}^(1/(p-1)) ] +2 とおけば、右辺 < ε.