Infinite Series ∑∞n=1Hnn32n\sum_{n=1}^\infty\frac{H_n}{n^32^n}

I’m trying to find a closed form for the following sum
n=1Hnn32n,
where Hn=nk=11k is a harmonic number.

Could you help me with it?

Answer

In the same spirit as Robert Israel’s answer and continuing Raymond Manzoni’s answer (both of them deserve the credit because of inspiring my answer) we have
n=1Hnxnn2=ζ(3)+12lnxln2(1x)+ln(1x)Li2(1x)+Li3(x)Li3(1x).
Dividing equation above by x and then integrating yields
n=1Hnxnn3=ζ(3)lnx+12lnxln2(1x)x dx+ln(1x)Li2(1x)x dx+Li4(x)Li3(1x)x dx.
Using IBP to evaluate the green integral by setting u=Li3(1x) and dv=1x dx, we obtain
Li3(1x)x dx=Li3(1x)lnx+lnxLi2(1x)1x dxx1x=Li3(1x)lnxln(1x)Li2(x)x dx.
Using Euler’s reflection formula for dilogarithm
Li2(x)+Li2(1x)=π26lnxln(1x),
then combining the blue integral in (1) and (2) yields
π26ln(1x)x dxlnxln2(1x)x dx=π26Li2(x)lnxln2(1x)x dx.
Setting x1x and using the identity Hn+1Hn=1n+1, the red integral becomes
lnxln2(1x)x dx=ln(1x)ln2x1x dx=n=1Hnxnln2x dx=n=1Hnxnln2x dx=n=1Hn2n2[xn dx]=n=1Hn2n2[xn+1n+1]=n=1Hn[xn+1ln2xn+12xn+1lnx(n+1)2+2xn+1(n+1)3]=ln2xn=1Hnxn+1n+12lnxn=1Hnxn+1(n+1)2+2n=1Hnxn+1(n+1)3=12ln2xln2(1x)2lnx[n=1Hn+1xn+1(n+1)2n=1xn+1(n+1)3]+2[n=1Hn+1xn+1(n+1)3n=1xn+1(n+1)4]=12ln2xln2(1x)2lnx[n=1Hnxnn2n=1xnn3]+2[n=1Hnxnn3n=1xnn4]=12ln2xln2(1x)2lnx[n=1Hnxnn2Li3(x)]+2[n=1Hnxnn3Li4(x)].
Putting all together, we have
n=1Hnxnn3=12ζ(3)lnx18ln2xln2(1x)+12lnx[n=1Hnxnn2Li3(x)]+Li4(x)π212Li2(x)12Li3(1x)lnx+C.
Setting x=1 to obtain the constant of integration,
n=1Hnn3=Li4(1)π212Li2(1)+Cπ472=π490π472+CC=π460.
Thus
n=1Hnxnn3=12ζ(3)lnx18ln2xln2(1x)+12lnx[n=1Hnxnn2Li3(x)]+Li4(x)π212Li2(x)12Li3(1x)lnx+π460.
Finally, setting x=12, we obtain
n=1Hn2nn3=π4720+ln4224ln28ζ(3)+Li4(12),
which matches Cleo’s answer.


References :

[1]  Harmonic number

[2]  Polylogarithm

Attribution
Source : Link , Question Author : OlegK , Answer Author : Community

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