how to find the following series: $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}$$ what i attempted was using symmetry like this \begin{align*} \sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)} &= 3 \sum_{i,j,n \ge 1} \frac n{n j i (n + j)(n + i)(j + i)} \\ &= 3 \sum_{i,j,n \ge 1} \frac1{i j (n + i)(n + j)(i + j)}. \end{align*} but i don't know how to proceed. also numerically evaluated $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\approx 2.47$$ which is roughly $$\frac32\sum_{n=1}^\infty \frac{H_n^2}{n^3}\approx2.47791418905674793594$$ so is $$\sum_{i,j,n\ge1}\frac{n + j + i}{n j i (n + j)(n + i)(j + i)}=\frac32\sum_{n=1}^\infty\frac{H_n^2}{n^3}?$$ how to prove this?
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- 5$\begingroup$ By symmetry, $$\begin{align*}S&=\frac{3}{2}\sum_{n,i,j\geq1}\frac{i+j}{nij(n+i)(n+j)(i+j)}\\&=\frac{3}{2}\sum_{n\geq1}\frac{1}{n}\left[\sum_{i\geq1}\frac{1}{i(n+i)}\right]^2\\&=\frac{3}{2}\sum_{n\geq1}\frac{1}{n}\left(\frac{H_n}{n}\right)^2.\end{align*}$$ I don't have a good idea how to evalute the last sum, though. $\endgroup$Sangchul Lee– Sangchul Lee2025-11-25 08:37:13 +00:00Commented 8 hours ago
- 1$\begingroup$ @SangchulLee math.stackexchange.com/questions/555266/… Maybe this is helpful. $\endgroup$Sahaj– Sahaj2025-11-25 08:58:08 +00:00Commented 8 hours ago
- 1$\begingroup$ @Sahaj, Thanks for the link! Now I remember I've seen that posting before. $\endgroup$Sangchul Lee– Sangchul Lee2025-11-25 09:02:21 +00:00Commented 8 hours ago
- 1$\begingroup$ @匚ㄖㄥᗪ乇ᗪ I Could you tell us why you believe the question you mentionned can help to prove the formula of the OP? $\endgroup$NN2– NN22025-11-25 09:28:05 +00:00Commented 7 hours ago
- 2$\begingroup$ @匚ㄖㄥᗪ乇ᗪ check the last sum with what in the question you mentionned? $\endgroup$NN2– NN22025-11-25 09:39:43 +00:00Commented 7 hours ago
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2 Answers
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2 I detail the first comment (in particular, the last step) of @SangchulLee $$\begin{align} S&:=\sum_{i,j,n \ge 1} \frac{n + j + i}{n j i (n + j)(n + i)(j + i)}\\ &= \frac{3}{2}\cdot\sum_{i,j,n \ge 1} \frac{j + i}{n j i (n + j)(n + i)(j + i)}\\ &= \frac{3}{2}\cdot\sum_{i,j,n \ge 1} \frac{1}{n j i (n + j)(n + i)}\\ &=\frac{3}{2}\cdot\sum_{i,j,n \ge 1} \frac{1}{n^3}\cdot\frac{n}{i(n+i)}\cdot\frac{n}{j(n+j)}\\ &=\frac{3}{2}\cdot\sum_{n \ge 1} \left(\frac{1}{n^3}\cdot\left(\sum_{i \ge 1}\left(\frac{1}{i}-\frac{1}{n+i}\right)\right)^2\right)\\ &=\frac{3}{2}\cdot\sum_{n \ge 1} \left(\frac{1}{n^3}\cdot\left(\sum_{i \ge 1}\frac{1}{i}-\sum_{i>n}\frac{1}{i}\right)^2\right) \hspace{0.5cm}\text{not very rigorous, but you have the idea}\\ &=\frac{3}{2}\cdot\sum_{n \ge 1} \left(\frac{1}{n^3}\cdot\left(\sum_{1\le i \le n}\frac{1}{i}\right)^2\right)\\ S&=\frac{3}{2}\cdot\sum_{n \ge 1} \left(\frac{H_n^2}{n^3}\right)\\ \end{align}$$ Q.E.D
- 1$\begingroup$ Thank you (and Sangchul). This is what I wanted $\endgroup$Wessel– Wessel2025-11-25 11:02:39 +00:00Commented 6 hours ago
- 1$\begingroup$ FYI, you can make the “not very rigorous step” rigorous using the summation by parts formula: en.wikipedia.org/wiki/Summation_by_parts. Cheers =) $\endgroup$David H– David H2025-11-25 15:56:02 +00:00Commented 1 hour ago
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derivation of closed form (following from NN2's); let $T=\frac23S$, then from Sangchul Lee's comment,$$ T=\sum_{n,i,j\ge1}\frac1{nij(n+i)(n+j)} $$by collapsing with respect to $j$ we get $$\sum_{i,n\ge1}\frac{H_n}{n^2i(n+i)}$$ which is mostly tautological, since $\sum_{i=1}^\infty\frac1{i(n+i)}=\frac{H_n}n$, returning to where we started.
however, by collapsing with respect to $n$ we get $$\sum_{i,j\ge1}\begin{cases}\frac{H_i-i\zeta(2,1+i)}{i^4}&i=j\\\frac{iH_j-jH_i}{i^2j^2(i-j)}&\mathrm{else}\end{cases}$$ which is useful! along the diagonal, $\sum_{i=1}^\infty\frac{H_i-i\zeta(2,1+i)}{i^4}=\sum_{i=1}^\infty\frac{H_i-i\zeta(2)+iH^{(2)}_i}{i^4}=\left(\sum_{i=1}^\infty\frac{H_i}{i^4}+\frac{H^{(2)}_i}{i^3}\right)-\zeta(2)\zeta(3)$, and off it we can fold $\sum\limits_{i,j\ge1\atop i\ne j}=2\sum\limits_{1\le i<j}$, from whence $2\sum_{i=1}^\infty\sum_{j=i+1}^\infty\frac{iH_j-jH_i}{i^2j^2(i-j)}=2\sum_{i=1}^\infty\frac{H_i^2}{i^3}+\sum_{j=i+1}^\infty\frac{iH_j}{i^2j^2(i-j)}$; $\sum_{i=1}^\infty\frac{H_i^2}{i^3}=T$ is what we want to begin with, so by subtracting both sides from $2T$, $$\begin{aligned} &T=\left(\sum_{i=1}^\infty\frac{H_i}{i^4}+\frac{H^{(2)}_i}{i^3}\right)-\zeta(2)\zeta(3)+2S+2\sum_{i=1}^\infty\sum_{j=i+1}^\infty\frac{iH_j}{i^2j^2(i-j)}\\ \iff&T=\zeta(2)\zeta(3)-\left(\sum_{i=1}^\infty\frac{H_i}{i^4}+\frac{H^{(2)}_i}{i^3}\right)+2\sum_{i=1}^\infty\sum_{j=i+1}^\infty\frac{iH_j}{i^2j^2(j-i)} \end{aligned}$$ then since $\frac1{i(j-i)}=\frac1j\left(\frac1i+\frac1{j-i}\right)$, the last part is $2\sum_{j=1}^\infty\frac{H_j}{j^2}\sum_{i=1}^{j-1}\frac1{i(j-i)}=4\sum_{j=1}^\infty\frac{H_jH_{j-1}}{j^3}=4\sum_{j=1}^\infty\frac{H_j^2}{j^3}-\frac{H_j}{j^4}$; the first addend of the summand is again $T$, so merging the two sums,$$\begin{aligned} &T=\zeta(2)\zeta(3)+4S-\sum_{i=1}^\infty5\frac{H_i}{i^4}+\frac{H^{(2)}_i}{i^3}\\ \iff&3S=\left(\sum_{i=1}^\infty5\frac{H_i}{i^4}+\frac{H^{(2)}_i}{i^3}\right)-\zeta(2)\zeta(3) \end{aligned}$$as a special case of this question (also explained quite concisely and elementarily in step 3 of this blogpost), $\sum_{i=1}^\infty\frac{H_i}{i^4}=3\zeta(5)-\zeta(2)\zeta(3)$, while (by ie. this question) $\sum_{i=1}^\infty\frac{H^{(2)}_i}{i^3}=3\zeta(2,3)+\zeta(2+3)$ which equals (by this question) $\zeta(2)\zeta(3)-\frac92\zeta(5)$, giving the final result $T=\frac72\zeta(5)-\zeta(2)\zeta(3)$, ie. $$S=\frac{21}4\zeta(5)-\frac32\zeta(2)\zeta(3)$$ edit: did not see Sahaj's comment linking this question, although that is closed and my answer does not quite trace out the same path as any answers to it