Series with summands involving harmonic numbers
Fuente:
arXiv
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| Autore principale: | |
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| Natura: | Preprint |
| Pubblicazione: |
2022
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| Accesso online: | |
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| _version_ | 1866917943438213120 |
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| author | Sun, Zhi-Wei |
| author_facet | Sun, Zhi-Wei |
| contents | For each positive integer $m$, the $m$th order harmonic numbers are given by $$H_n^{(m)}=\sum_{0<k\le n}\frac1{k^m}\ \ (n=0,1,2,\ldots).$$ We discover exact values of some series involving harmonic numbers of order not exceeding four. For example, we conjecture that $$\sum_{k=0}^\infty(6k+1)\frac{\binom{2k}k^3}{256^k}\left(H_{2k}^{(3)}-\frac{7}{64}H_{k}^{(3)}\right)
=\frac{25ζ(3)}{8π}-G,$$ where $G$ denotes the Catalan constant $\sum_{k=0}^\infty(-1)^k/(2k+1)^2$. This paper contains $70$ conjectures posed by the author during 2022--2023. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2210_07238 |
| institution | arXiv |
| publishDate | 2022 |
| record_format | arxiv |
| spellingShingle | Series with summands involving harmonic numbers Sun, Zhi-Wei Number Theory 11B65, 05A19, 11A07, 11B68 For each positive integer $m$, the $m$th order harmonic numbers are given by $$H_n^{(m)}=\sum_{0<k\le n}\frac1{k^m}\ \ (n=0,1,2,\ldots).$$ We discover exact values of some series involving harmonic numbers of order not exceeding four. For example, we conjecture that $$\sum_{k=0}^\infty(6k+1)\frac{\binom{2k}k^3}{256^k}\left(H_{2k}^{(3)}-\frac{7}{64}H_{k}^{(3)}\right) =\frac{25ζ(3)}{8π}-G,$$ where $G$ denotes the Catalan constant $\sum_{k=0}^\infty(-1)^k/(2k+1)^2$. This paper contains $70$ conjectures posed by the author during 2022--2023. |
| title | Series with summands involving harmonic numbers |
| topic | Number Theory 11B65, 05A19, 11A07, 11B68 |
| url | https://arxiv.org/abs/2210.07238 |