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I am looking for the closed form of:

$$\sum_{n=0}^{\infty}H_{n+2}^2\sum_{k=0}^{n}(-1)^k{ n \choose k}\frac{1}{k+j}=F(j);\quad j\ge2$$

Where $H_n$ is Harmonic numbers and $\zeta(s)$ is Reimann zeta function

I have managed to figure out the first three values for $j=2,3$ and $4$,

$$F(2)=1+\zeta(2)+H_1+1$$ $$F(3)=\zeta(2)-3\zeta(3)+H_2+2$$ $$F(4)=2\zeta(2)-6\zeta(3)+H_3+3$$ $$F(j)=?$$

metamorphy
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Sibawayh
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    One maybe nonobvious simplification you can do is $$F(j) = \frac{1}{j} \sum_n \frac{H_{n+2}^2}{\binom{n+j}{n}}$$ (I got this by copying your definition into sage, so you should double check it) – HallaSurvivor May 10 '21 at 16:56

1 Answers1

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From Vandermonde's identity, $$ \sum_{k=0}^n(-1)^k\binom nk\frac{1}{k+j}=\frac 1j\frac 1{\binom{n+j}j} $$ Hence \begin{align} F(j)&=\sum_{0\leq n}H_{n+2}^2\sum_{k=0}^n(-1)^k\binom nk\frac 1{k+j}\\ &=\frac 1j\sum_{0\leq n}\frac{H_{n+2}^2}{\binom{n+j}j}\\ &=\sum_{0<n}\frac{H_{n+1}^2}{n\binom{n+j-1}{j-1}}\\ &=\sum_{0<n}\frac{\left(H_{n}+\frac 1{n+1}\right)^2}{n\binom{n+j-1}{j-1}}\\ &=\sum_{0<n}\frac{H_n^2}{n\binom{n+j-1}{j-1}}+2\sum_{0<n}\frac{H_n}{n(n+1)\binom{n+j-1}{j-1}}+\sum_{0<n}\frac{1}{n(n+1)^2\binom{n+j-1}{j-1}} \end{align}

Define $S_m, H_m$ and $\zeta^m$ as follows, \begin{align} S_m(k_1,\dots,k_r)&:=\sum_{0<n_1<\cdots<n_r}\frac 1{n_1^{k_1}\cdots n_r^{k_r}\binom{n_r+m}{n_r}}\\ H_m(k_1,\dots,k_r)&:=\sum_{0<n_1<\cdots<n_r\leq m}\frac 1{n_1^{k_1}\cdots n_r^{k_r}}\\ \zeta^m(k_1,\dots,k_r)&:=\sum_{m\leq n_1<\cdots<n_r}\frac{1}{n_1^{k_1}\cdots n_r^{k_r}} \end{align} According to connected sum method (https://arxiv.org/abs/2006.09076, Proposition 5.1)

Let $k_r\geq 2$ and $(l_1,\dots,l_s)$ be dual index of $(k_1,\dots,k_r)$, then $$ S_m(k_1,\dots,k_r,\{1\}^t)=\frac 1{m^t}\sum_{m<n_1<\cdots<n_s}\frac 1{n_1^{l_1}\cdots n_s^{l_s}}=\frac 1{m^t}\zeta^{m+1}(l_1,\dots,l_s) $$ by the following identity. $$ \sum_{n<n_1<\cdots<n_t}\frac 1{n_1\cdots n_t\binom{n_t+m}{n_t}}=\frac 1{m^t\binom{n+m}{n}} $$

Therefore, first term is \begin{align} \sum_{0<n}\frac{H_n^2}{n\binom{n+j-1}{j-1}}&=\sum_{0<l,m\leq n}\frac{1}{lmn\binom{n+j-1}{j-1}}\\ &=S_{j-1}(3)+2S_{j-1}(1,2)+S_{j-1}(2,1)+2S_{j-1}(1,1,1)\\ &=\zeta^j(1,2)+2\zeta^j(3)+\frac 1{j-1}\zeta^j(2)+\frac{2}{(j-1)^3} \end{align} Second term is \begin{align} 2\sum_{0<n}\frac{H_n}{n(n+1)\binom{n+j-1}{j-1}}&=2\sum_{0<n}\frac{H_n}{n\binom{n+j-1}{j-1}}-2\sum_{0<n}\frac{H_n}{(n+1)\binom{n+j-1}{j-1}}\\ &=2S_{j-1}(1,1)+2S_{j-1}(2)-2(j-1)\sum_{0<n}\frac{H_{n-1}}{n^2\binom{n+j-2}{j-2}}\\ &=\frac{2}{(j-1)^2}+2\zeta^j(2)-2(j-1)S_{j-2}(1,2)\\ &=\frac{2}{(j-1)^2}+2\zeta^j(2)-2(j-1)\zeta^{j-1}(3)\\ &=2\zeta^j(2)-2(j-1)\zeta^j(3) \end{align} Third term is \begin{align} \sum_{0<n}\frac 1{n(n+1)^2\binom{n+j-1}{j-1}}&=\sum_{0<n}\frac 1{n\binom{n+j-1}{j-1}}-\sum_{0<n}\frac 1{(n+1)\binom{n+j-1}{j-1}}-\sum_{0<n}\frac 1{(n+1)^2\binom{n+j-1}{j-1}}\\ &=S_{j-1}(1)-(j-1)\left(\sum_{0<n}\frac 1{n^2\binom{n+j-2}{j-2}}+\sum_{0<n}\frac 1{n^3\binom{n+j-2}{j-2}}\right)+2\\ &=S_{j-1}(1)-(j-1)(S_{j-2}(2)+S_{j-2}(3))+2\\ &=\frac 1{j-1}-(j-1)(\zeta^{j-1}(2)+\zeta^{j-1}(1,2))+2\\ &=\frac 1{j-1}-(j-1)\left(\frac 1{(j-1)^2}+\left(1+\frac{1}{j-1}\right)\zeta^j(2)+\zeta^j(1,2)\right)+2\\ &=-j\zeta^j(2)-(j-1)\zeta^j(1,2)+2 \end{align} Hence, \begin{align} F(j)&=\left(\zeta^j(1,2)+2\zeta^j(3)+\frac 1{j-1}\zeta^j(2)+\frac{2}{(j-1)^3}\right)+\left(2\zeta^j(2)-2(j-1)\zeta^j(3)\right)\\ &+\left(-j\zeta^j(2)-(j-1)\zeta^j(1,2)+2\right)\\ &=(2-j)\zeta^j(1,2)+2(2-j)\zeta^j(3)+\left(2+\frac{1}{j-1}-j\right)\zeta^j(2)+\frac 2{(j-1)^3}+2 \end{align} Finally, we use \begin{align} \zeta^j(k)&=\zeta(k)-H_{j-1}(k)\\ \zeta^j(1,2)&=\sum_{0<n<m}\frac 1{nm^2}-\sum_{\substack{0<n<j\\n<m}}\frac 1{nm^2}\\ &=\zeta(1,2)-\sum_{0<n<j\leq m}\frac 1{nm^2}-\sum_{0<n<m<j}\frac 1{nm^2}\\ &=\zeta(3)-H_{j-1}\zeta^j(2)-H_{j-1}(1,2)\\ &=\zeta(3)-H_{j-1}(\zeta(2)-H_{j-1}(2))-H_{j-1}(1,2) \end{align} Hence, \begin{align} F(j)&=(2-j)(\zeta(3)-H_{j-1}(\zeta(2)-H_{j-1}(2))-H_{j-1}(1,2))\\ &+2(2-j)(\zeta(3)-H_{j-1}(3))+\left(2+\frac{1}{j-1}-j\right)(\zeta(2)-H_{j-1}(2))+\frac 2{(j-1)^3}+2\\ &=(2-j)(3\zeta(3)-2H_{j-1}(3))+\left(2+\frac{1}{j-1}-j-(2-j)H_{j-1}\right)(\zeta(2)-H_{j-1}(2))\\ &+\frac 2{(j-1)^3}+2-(2-j)H_{j-1}(1,2)\\ &=(2-j)(3\zeta(3)-2H_{j-1}(3))+\left(3-j+(j-2)H_{j-2}\right)(\zeta(2)-H_{j-1}(2))\\ &+\frac 2{(j-1)^3}+2+(j-2)H_{j-1}(1,2)\\ \end{align}