Infinite Series ∞∑n=1(Hn)2n3\sum\limits_{n=1}^\infty\frac{(H_n)^2}{n^3}

How to prove that
n=1(Hn)2n3=72ζ(5)ζ(2)ζ(3)
Hn denotes the harmonic numbers.

Answer

Identities such as this can be proved with the help of Cauchy’s residue theorem. If f is a meromorphic function such that |f(z)|=o(z1) as |z| on a sequence of concentric circles about the origin then, the residue theorem gives
aRes(f,a)=0.(1)
Here, the sum is over all poles of f and Res(f,a) is the residue of f at a. The tricky part is finding the right function f. Flajolet & Salvy1 show how to prove a whole set of identities of this form. For example (all sums are over n=1 to ),
Hnn2=2ζ(3),Hnn3=54ζ(4),Hnn4=3ζ(5)ζ(2)ζ(3),(Hn)2n2=174ζ(4),(Hn)2n3=72ζ(5)ζ(2)ζ(3),(2)(Hn)2n4=9724ζ(6)2ζ(3)2,(Hn)3n4=23116ζ(7)514ζ(3)ζ(4)+2ζ(2)ζ(5)
We also have the set of identities due to Euler (for q2)
n=1Hnnq=(1+q2)ζ(q+1)12q2k=1ζ(k+1)ζ(qk).
It is mentioned in Flajolet & Salvy that identities of this form do not always exist and, in particular, there is unlikely to be any finite formula for (Hn)3/nq in terms of zeta values when q is an odd number exceeding 10.

I’ll give the function f which generates the identity (2) asked for, following Flajolet & Salvy (specializing to this example).
Let ψ be the digamma function
ψ(z)=ddzlogΓ(z)=γ1z+n=1(1n1n+z).
This is bounded by O(|z|ϵ) on circles of radius n+1/2 about the origin and has poles at the nonnegative integers. To prove the required identity (2), it is easiest to break this down into three identities (although, you could add the three choices of f below and do it in one go).


Taking the function f(z)=13z3(ψ(z)+γ)3, this has poles at the nonnegative integers. Expand each term about 0 and positive integers n>0,
ψ(z)+γ=1zζ(2)zζ(3)z2ζ(4)z3ζ(5)z4+O(z5)ψ(nz)+γ=1z+Hn(H(2)n+ζ(2))z+O(z2)(n+z)3=n33n4z+6n5z2+O(z3)
Here, H(2)n is the generalized harmonic number knk2. Multiplying the terms together and extracting the coefficients of z1 gives the residues of f.
Res(f,0)=2ζ(2)ζ(3)ζ(5),Res(f,n)=n3(H2nH(2)nζ(2))3n4Hn+2n5.
Summing over n and applying the residue theorem,
H2nn3H(2)nn33Hnn4+ζ(5)+ζ(2)ζ(3)=0.(3)
Now, take the function f(z)=12z4(ψ(z)+γ)2. This again has poles at the nonnegative integers. Using the expansions above together with
(n+z)4=n44n5z+O(z2)
we can compute the residues as before
Res(f,0)=ζ(2)ζ(3)ζ(5),Res(f,n)=n4Hn2n5.
Applying the residue theorem again gives
Hnn43ζ(5)+ζ(2)ζ(3)=0.(4)
Finally take f(z)=12πz3cot(πz)ψ(z) and use the expansions
ψ(z)=z2+ζ(2)+2ζ(3)z+3ζ(4)z2+4ζ(5)z3+O(z4)ψ(nz)=z2+H(2)n+ζ(2)+O(z)ψ(nz)=ζ(2)+n2H(2)n+O(z)πcot(π(±n+z))=z113π2z+cz3+O(z5)=z12ζ(2)z+cz3+O(z5)
(some constant c) to compute the residues
Res(f,0)=2ζ(5)2ζ(2)ζ(3),Res(f,n)=12n3(H(2)nζ(2))+3n5,Res(f,n)=12n3(ζ(2)+n2H(2)n).
Summing over n and applying the residue theorem,
H(2)nn3+92ζ(5)3ζ(2)ζ(3)=0.(5)
Adding identities (3), 3 times (4), and (5) gives the required result.


I’ll just add a note that the use of the digamma function, cotangent and residue theorem above are not really required. It has been mentioned in Noam D. Elkie’s answer that such results can be proved by elementary, but clever, algebraic manipulations. Applied to rational functions, the residue theorem gives algebraic identities which can be easily verified. Also, the digamma function and cotangent can be expressed as sums over terms of the form i1(i+z)1 over integer i. So, expanding the functions f above as infinite sums over rational functions before applying the residue theorem reduces the argument to one involving summing over elementary identities.
In particular, applying the residue theorem to the functions 1z2(1i1iz)(1j1jz), 1z3(1i1i+z)1(jz)2 and 1z3(1i1iz)(1j1jz)(1k1kz) gives, respectively,
1i4(1j1ji)+1j4(1i1ij)+1i2j3+1i3j2=0,3j4(1i1i+j)1j3(i+j)21i3(i+j)22i2j3+1i3j2=0,(ijk)1i3(1j1ji)(1k1ki)=0.
In the last identity, the summation refers to the sum over the three cyclic permutations of i,j,k. Summing these identities over positive integers i,j,k and cancelling terms of the form 1i and 1i±j leads to identities (3,4,5) above.

1 Euler sums and contour integral representations, P. Flajolet, B. Salvy, Experimental Mathematics Volume 7, Issue 1 (1998), 15-35. (link)

Attribution
Source : Link , Question Author : Community , Answer Author : George Lowther

Leave a Comment