The result is
1/4 (ArcCosh[3] ArcSinh[1] -
ArcSinh[1 - Sqrt[2]] Log[7 - 4 Sqrt[2] - 2 Sqrt[2 (10 - 7 Sqrt[2])]])
You can get to it using the integral representation of LegendreQ and then
pulling the integral before the sum.
With this integral repesentation of LegendreQ (omitting a purely imaginary part)
I1=Integrate[(1/Sqrt[2] + I Cosh[t]/Sqrt[2])^(-1 - n), {t, 0, Infinity}]
you get
Sum[1/(n+1)I1,{n,0,Infinity}],
then exchange summation and integral to arrive at
Integrate[(-I (-I +
Cosh[t]) Log[(-I + I Sqrt[2] + Cosh[t])/(-I + Cosh[t])]/(Sqrt[
2] (1/Sqrt[2] + (I Cosh[t])/Sqrt[2]))), {t, 0, Infinity}]
Now evaluate the antiderivative, enter the limits and take the real part of
the result. By checking the remaining PolyLogs you find they are all zero.
For (-1<z<1) I get the more general result
Re[PolyLog[2, 2 /(1 - Sqrt[I Sqrt[1 - z^2] - z])] + PolyLog[2, 2 /(1 + Sqrt[I Sqrt[1 - z^2] - z])] - PolyLog[2, 2 /(1 - Sqrt[2 z (z - I Sqrt[1 - z^2]) - 1])] - PolyLog[2, 2 /(1 + Sqrt[2 z (z - I Sqrt[1 - z^2]) - 1])]]
The real part of the above is
ArcTanh[z]^2/2 + 1/2 Log[(Sqrt[2] - Sqrt[1 - z])/Sqrt[1 + z]]^2 + ArcTanh[Sqrt[1 - z]/Sqrt[2]] Log[2] +1/2 (ArcTanh[Sqrt[1 - z]/Sqrt[2]] +Log[2]) Log[(1 + z)/(3 + 2 Sqrt[2 - 2 z] - z)] + \[Pi]^2/24 + 1/2 Log[1 + Sqrt[2]]^2 + 1/4 PolyLog[2, -3 - 2 Sqrt[2]] + 1/4 PolyLog[2, -3 + 2 Sqrt[2]]
The derivation is along the line of the special case, I integrated only the real part of the integral representation to obtain a real expression.
We can even get rid of the PolyLog constants:
ArcTanh[z]^2/2 + ArcTanh[Sqrt[1 - z]/Sqrt[2]] Log[2] + 1/2 Log[(Sqrt[2] -
Sqrt[1 - z])/Sqrt[1 + z]]^2 + 1/2 (ArcTanh[Sqrt[1 - z]/Sqrt[2]] + Log[2]) Log[(1 + z)/(3 + 2 Sqrt[2 - 2 z] - z)] - 4 ArcCoth[Sqrt[2]]^2 +
4 ArcSinh[1]^2 + 2 ArcSinh[1 - Sqrt[2]] Log[7 - 4 Sqrt[2] - 2 Sqrt[2] Sqrt[10 - 7 Sqrt[2]]] - Log[7 - 4 Sqrt[2] + 2 Sqrt[2] Sqrt[10 - 7 Sqrt[2]]]^2
I noticed that this can further be simplified to a short and beautiful
ArcTanh[z]^2/2-ArcTanh[Sqrt[1 - z]/Sqrt[2]]^2 +
1/2 Log[(Sqrt[2] - Sqrt[1 - z])/Sqrt[1 + z]]^2.
And still shorter, so that we finally have:
Sum[LegendreQ[n, z]/(n + 1), {n, 0, Infinity}] =
(ArcTanh[z]^2 - Log[(Sqrt[2] + Sqrt[1 - z])/Sqrt[1 + z]]^2)/2, (-1 < z < 1).
NSum[RealAbs[LegendreQ[n, Sqrt[2]/2]]/(n + 1), {n, 0, Infinity}, Method -> {"WynnEpsilon", "ExtraTerms" -> 20, "Degree" -> 1}, NSumTerms -> 20, WorkingPrecision -> 30]
is running without any respose for dozen minutes. $\endgroup$NSum
's option settingVerifyConvergence -> False
that computation runs for less than 0.03 on my laptop with Version 12.1. $\endgroup$