Euler Sums系列(五)

\[\Large\displaystyle \sum_{n=1}^{\infty} \frac{\widetilde{H_n}}{n^{3}} \]

where \(\widetilde{H_n}\) is the alternating harmonic number.


\(\Large\mathbf{Solution:}\)
Namely,

\[\widetilde{H_n} = \ln (2) + (-1)^{n-1} \int_{0}^{1} \frac{x^{n}}{1+x} \mathrm dx \]

Using that representation,

\[\begin{align*} {\sum_{n=1}^{\infty} \frac{\widetilde{H_n}}{n^{3}}} &= \ln (2) \sum_{n=1}^{\infty} \frac{1}{n^{3}} + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{3}} \int_{0}^{1} \frac{x^{n}}{1+x} \mathrm dx \\ &= \zeta(3) \ln(2) - \int_{0}^{1} \frac{1}{1+x} \sum_{n=1}^{\infty} \frac{(-x)^{n}}{n^{3}}\mathrm dx \\ &= \zeta(3) \ln(2) - \int_{0}^{1} \frac{\text{Li}_{3}(-x)}{1+x} \mathrm dx \\ &= \zeta(3) \ln(2) - \text{Li}_{3}(-x) \ln(1+x) \Bigg|^{1}_{0} + \int_{0}^{1} \frac{\text{Li}_{2}(-x) \ln(1+x)}{x} \mathrm dx \\ &= \zeta(3) \ln(2) + \frac{3}{4} \zeta(3) \ln(2) - \frac{1}{2} \Big( \text{Li}_{2}(-1) \Big)^{2} \\ &= \Large\boxed{\displaystyle \color{blue}{\frac{7}{4} \zeta(3) \ln(2) - \frac{\pi^{4}}{288}}} \end{align*} \]

This also can be Evaluated by using the fact that

\[\large\boxed{\displaystyle \color{DarkOrange} {\sum_{n=1}^\infty \frac{\widetilde{H_n}}{n^q} = \zeta(q)\ln(2)-\frac{q}{2}\zeta(q+1)+2\eta(z)+\sum_{k=1}^q \eta(k)\eta(q-k+1)}} \]

where \(\eta(z)\) is the Dirichlet Eta Function and \(\displaystyle \widetilde{H_n}=\sum_{j=1}^n \frac{(-1)^{j-1}}{j}\).

posted @ 2016-05-12 16:09  Renascence_5  阅读(416)  评论(0编辑  收藏