Question

k趋于无穷,求级数k(k+1)/（2^k+1）的和

Answer 1

做幂级数

f(x) = Σ(k>=1)[k(k+1)(x/2)^(k-1)],

利用逐项积分定理,得

g(x) = ∫[0,x]f(t)dt

= Σ(k>=1)k(k+1)∫[0,x][(t/2)^(k-1)]dt

= 2Σ(k>=1)[(k+1)(t/2)^k],

再积分

∫[0,x]g(t)dt

= Σ(k>=1)(k+1)∫[0,x][(t/2)^k]dt

= 4Σ(k>=1)[(t/2)^(k+1)]

= 4{1/[1-(x/2)] - 1 - x/2},

（接下来,两次求导,再取 x=1,即可得）……