[Largedisplaystyle sum_{n=1}^{infty} frac{widetilde{H_n}}{n^{3}}
]
where (widetilde{H_n}) is the alternating harmonic number.
(Largemathbf{Solution:})
Namely,
[widetilde{H_n} = ln (2) + (-1)^{n-1} int_{0}^{1} frac{x^{n}}{1+x} mathrm dx
]
Using that representation,
[egin{align*} {sum_{n=1}^{infty} frac{widetilde{H_n}}{n^{3}}} &= ln (2) sum_{n=1}^{infty} frac{1}{n^{3}} + sum_{n=1}^{infty} frac{(-1)^{n-1}}{n^{3}} int_{0}^{1} frac{x^{n}}{1+x} mathrm dx \ &= zeta(3) ln(2) - int_{0}^{1} frac{1}{1+x} sum_{n=1}^{infty} frac{(-x)^{n}}{n^{3}}mathrm dx \ &= zeta(3) ln(2) - int_{0}^{1} frac{ ext{Li}_{3}(-x)}{1+x} mathrm dx \ &= zeta(3) ln(2) - ext{Li}_{3}(-x) ln(1+x) Bigg|^{1}_{0} + int_{0}^{1} frac{ ext{Li}_{2}(-x) ln(1+x)}{x} mathrm dx \ &= zeta(3) ln(2) + frac{3}{4} zeta(3) ln(2) - frac{1}{2} Big( ext{Li}_{2}(-1) Big)^{2} \ &= Largeoxed{displaystyle color{blue}{frac{7}{4} zeta(3) ln(2) - frac{pi^{4}}{288}}} end{align*}
]
This also can be Evaluated by using the fact that
[largeoxed{displaystyle color{DarkOrange} {sum_{n=1}^infty frac{widetilde{H_n}}{n^q} = zeta(q)ln(2)-frac{q}{2}zeta(q+1)+2eta(z)+sum_{k=1}^q eta(k)eta(q-k+1)}}
]
where (eta(z)) is the Dirichlet Eta Function and (displaystyle widetilde{H_n}=sum_{j=1}^n frac{(-1)^{j-1}}{j}).