Question Number 100237 by mathmax by abdo last updated on 25/Jun/20

calculate Σ_(k=0) ^n  (((−1)^k )/((k+1)^3 ))C_n ^k

Answered by maths mind last updated on 26/Jun/20

x^a (1−x)^n =Σx^a C_n ^k (−x)^k =ΣC_n ^k (−1)^k x^(a+k)   ⇒∫_0 ^1 x^a (1−x)^n dx=Σ_(k=0) ^n (((−1)^k C_n ^k )/(a+k+1))...E  β(a,b)=∫_0 ^1 x^(a−1) (1−x)^(b−1) dx  E=β(a+1,n+1)=Σ_(k=0) ^n (((−1)^k C_n ^k )/(a+k+1))  ⇒(∂^2 /∂a^2 )β(a+1,n+1)=Σ_(k=0) ^n ((2(−1)^k C_n ^k )/((a+k+1)^3 ))  ⇔  (∂^2 /∂a^2 )β(a+1,n+1)∣_(a=0) =Σ_(k=0) ^n ((2(−1)^k C_n ^k )/((k+1)^3 ))  ((∂β(a+1,n+1))/∂a)=β(a+1,n+1)(Ψ(a+1)−Ψ(n+a+2))  ⇒β(a+1,n+1)(Ψ(a+1)−Ψ(n+a+2))^2 +β(a+1,n+1)(Ψ′(a+1)−Ψ′(n+a+2))  ⇔  Σ2(((−1)^k )/((k+1)^3 ))C_n ^k =β(1,n+1)(Ψ(1)−Ψ(n+2))+β(1,n+1)(Ψ′(1)−Ψ′(n+2))  β(1,n+1)=(1/(n+1))  Ψ(1)−Ψ(n+2)=−H_(n+1)   Ψ′(z)=ζ(2,z)=Σ_(k=0) ^∞ (1/((z+k)^2 ))  Ψ′(1)−Ψ′(n+2)=−H_(n+1) ^((2))   we get  =(1/(n+1))(−H_(n+1) −H_(n+1) ^((2)) )=−(1/(n+1))(H_(n+1) +H_(n+1) ^((2)) )=Σ((2(−1)^k C_n ^k )/((k+1)^3 ))  ⇒−(1/(2(n+1)))(H_(n+1) +H_(n+1) ^((2)) )=Σ_(k=0) ^n (((−1)^k C_n ^k )/((k+1)^3 ))

Commented bymathmax by abdo last updated on 26/Jun/20

thank you sir.