Euler Sums系列（四）

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Euler Sums系列（四）

\[\Large\displaystyle \sum_{n=1}^\infty (-1)^n \frac{H_n}{2n+1}=\mathbf{G}-\frac{\pi}{2}\ln(2)\] \(\Large\mathbf{Proof:}\) \(\Large\mathbf{Method~One:}\) Using the relation \(\displaystyle H_{n} = \int_{0}^{1} \frac{1-x^n}{1-x} \mathrm{d}x\), we find…

Euler Sums系列（六）

\[\Large\displaystyle \sum_{n=1}^{\infty}\frac{H_{2n}}{n(6n+1)}\] \(\Large\mathbf{Solution:}\) Let \(S\) denote the sum. Then \[\begin{align*} S=\sum_{n=1}^\infty \frac{H_{2n}}{n(6n+1)} &= \sum_{n=1}^\infty\frac{H_{2n}}{n}\int_0^1 x^{6n}\mathrm dx \\…

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} \math…

Euler Sums系列（一）

\[\Large\sum_{n=1}^{\infty} \frac{H_{n}}{2^nn^4}\] \(\Large\mathbf{Solution:}\) Let \[\mathcal{S}=\sum^\infty_{n=1}\frac{H_n}{n^42^n}\] We first consider a slightly different yet related sum. The main idea is to solve this sum with two different meth…

Euler Sums系列（三）

\[\Large\sum_{n=1}^{\infty}\frac{\left(H_{n}^{(2)}\right)^{2}}{n^{2}}=\frac{19}{24}\zeta(6)+\zeta^{2}(3)\] \(\Large\mathbf{Proof:}\) We use the Abel's rearrangement over the \(N\)-th partial sum of the series, \[\begin{align*}\sum\limits_{n=1}^{N}\fr…

Euler Sums系列（二）

\[\Large\sum_{n=0}^\infty \frac{H_{2n+1}}{(2n+1)^2}=\frac{21}{16}\zeta(3)\] \(\Large\mathbf{Proof:}\) Let \(\displaystyle S_1=\sum_{n=1}^\infty \frac{H_n}{n^2}\) and \(\displaystyle S_2 = \sum_{n=1}^\infty(-1)^{n+1}\frac{H_n}{n^2}\). Then, our sum ca…