LI YINGYING In= ∑、/(a dt+ k=n+1 t)k(k+1) k=n+1A(k+1) t(1-t) k=n+ k(k+1)(k+t Then rn=ri(n)+ n+1 k(k+1)(k+t) 1 1 t(1-t) k(k+1)(k+t)k(k+1)(k+2) t(1-t) k(k+1)(k+2) t(1-t)(2-t) k(k+1)(k+2)(k+t) 2 ta-ddt(a k=n+ k(k+1)(k+1)(k+2) dt+ k(k+1)(k+ + n+1)(n+2) Let r2(n)=∑ t(1-t)(2-t) k(k+1)(k+2)(k+ (7) n=r2(n)+a1 n+1(n+1)(n+2) Define for n∈N,m≥2, rm(m)=∑ t(1-t)(2-t)…(m-t) k(k+1)(k+2)…(k+m)(k+t) dt, am t(1-t)…(m-1-t)dt.(9)2 LI YINGYING From (∗) we obtain rn = X∞ k=n+1 Z 1 0 t k(k + t) dt = X∞ k=n+1 Z 1 0 t µ 1 k(k + t) − 1 k(k + 1)¶ dt + Z 1 0 tdt X∞ k=n+1 1 k(k + 1) = X∞ k=n+1 Z 1 0 t(1 − t) k(k + 1)(k + t) dt + 1 n + 1 Z 1 0 tdt. Write r1(n) = X∞ k=n+1 Z 1 0 t(1 − t) k(k + 1)(k + t) dt, a1 = Z 1 0 tdt. (5) Then rn = r1(n) + a1 n + 1 . (6) Moreover, r1(n) = X∞ k=n+1 Z 1 0 t(1 − t) k(k + 1)(k + t) dt = X∞ k=n+1 Z 1 0 t(1 − t) µ 1 k(k + 1)(k + t) − 1 k(k + 1)(k + 2)¶ dt + X∞ k=n+1 Z 1 0 t(1 − t) k(k + 1)(k + 2)dt = X∞ k=n+1 Z 1 0 t(1 − t)(2 − t) k(k + 1)(k + 2)(k + t) dt + X∞ k=n+1 1 2 Z 1 0 t(1 − t) dt µ 1 k(k + 1) − 1 (k + 1)(k + 2)¶ = X∞ k=n+1 Z 1 0 t(1 − t)(2 − t) k(k + 1)(k + 2)(k + t) dt + 1 2 Z 1 0 t(1 − t) dt 1 (n + 1)(n + 2). Let r2(n) = X∞ k=n+1 Z 1 0 t(1 − t)(2 − t) k(k + 1)(k + 2)(k + t) dt, a2 = 1 2 Z 1 0 t(1 − t)dt. (7) Then rn = r2(n) + a1 n + 1 + a2 (n + 1)(n + 2). (8) Define for m ∈ N, m ≥ 2, rm(n) = X∞ k=n+1 Z 1 0 t(1 − t)(2 − t)· · ·(m − t) k(k + 1)(k + 2)· · ·(k + m)(k + t) dt, am = 1 m Z 1 0 t(1 − t)· · ·(m − 1 − t) dt. (9)