I’m trying to find a closed form for the following sum

∞∑n=1Hnn32n,

where Hn=n∑k=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(1−x)+ln(1−x)Li2(1−x)+Li3(x)−Li3(1−x).

Dividing equation above by x and then integrating yields

∞∑n=1Hnxnn3=ζ(3)lnx+12∫lnxln2(1−x)x dx+∫ln(1−x)Li2(1−x)x dx+Li4(x)−∫Li3(1−x)x dx.

Using IBP to evaluate the green integral by setting u=Li3(1−x) and dv=1x dx, we obtain

∫Li3(1−x)x dx=Li3(1−x)lnx+∫lnxLi2(1−x)1−x dxx↦1−x=Li3(1−x)lnx−∫ln(1−x)Li2(x)x dx.

Using Euler’s reflection formula for dilogarithm

Li2(x)+Li2(1−x)=π26−lnxln(1−x),

then combining the blue integral in (1) and (2) yields

π26∫ln(1−x)x dx−∫lnxln2(1−x)x dx=−π26Li2(x)−∫lnxln2(1−x)x dx.

Setting x↦1−x and using the identity Hn+1−Hn=1n+1, the red integral becomes

∫lnxln2(1−x)x dx=−∫ln(1−x)ln2x1−x dx=∫∞∑n=1Hnxnln2x dx=∞∑n=1Hn∫xnln2x dx=∞∑n=1Hn∂2∂n2[∫xn dx]=∞∑n=1Hn∂2∂n2[xn+1n+1]=∞∑n=1Hn[xn+1ln2xn+1−2xn+1lnx(n+1)2+2xn+1(n+1)3]=ln2x∞∑n=1Hnxn+1n+1−2lnx∞∑n=1Hnxn+1(n+1)2+2∞∑n=1Hnxn+1(n+1)3=12ln2xln2(1−x)−2lnx[∞∑n=1Hn+1xn+1(n+1)2−∞∑n=1xn+1(n+1)3]+2[∞∑n=1Hn+1xn+1(n+1)3−∞∑n=1xn+1(n+1)4]=12ln2xln2(1−x)−2lnx[∞∑n=1Hnxnn2−∞∑n=1xnn3]+2[∞∑n=1Hnxnn3−∞∑n=1xnn4]=12ln2xln2(1−x)−2lnx[∞∑n=1Hnxnn2−Li3(x)]+2[∞∑n=1Hnxnn3−Li4(x)].

Putting all together, we have

∞∑n=1Hnxnn3=12ζ(3)lnx−18ln2xln2(1−x)+12lnx[∞∑n=1Hnxnn2−Li3(x)]+Li4(x)−π212Li2(x)−12Li3(1−x)lnx+C.

Setting x=1 to obtain the constant of integration,

∞∑n=1Hnn3=Li4(1)−π212Li2(1)+Cπ472=π490−π472+CC=π460.

Thus

∞∑n=1Hnxnn3=12ζ(3)lnx−18ln2xln2(1−x)+12lnx[∞∑n=1Hnxnn2−Li3(x)]+Li4(x)−π212Li2(x)−12Li3(1−x)lnx+π460.

Finally, setting x=12, we obtain

∞∑n=1Hn2nn3=π4720+ln4224−ln28ζ(3)+Li4(12),

which matches Cleo’s answer.

**References :**

[1] Harmonic number

[2] Polylogarithm

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