Find triple summation rel in a closed form $S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$

Question

Evaluate $\displaystyle S=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$

My attempt :

Let $$A=\sum_{k=1}^{m}\frac{1}{k(k+1)} =\sum_{k=1}^{m}\left( \frac1{k}-\frac1{k+1} \right) = \frac{m}{m+1}$$

and a second sum :

$$B=\sum_{m=1}^{n}\frac{1}{(m+1)^{2}}$$

from here how I can complete ??

Answer 1

Continuing from the calculation above we have to sum:

$$S=\sum_{n=1}^\infty\frac{1}{n(n+1)}\sum_{m=1}^n\frac{1}{(m+1)^2}=\sum_{n=1}^{\infty}\frac{H_2(n+1)-1}{n(n+1)}\\=\sum_{n=1}^{\infty}[\frac{H_2(n+1)}{n}-\frac{H_2(n+1)}{n+1}]-\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$$

Using the identity $H_2(n+1)=H_2(n)+\frac{1}{(n+1)^2}$

$$S=\sum_{n=1}^{\infty}[\frac{H_2(n)}{n}-\frac{H_2(n+1)}{n+1}]-1+\sum_{n=1}^{\infty}\frac{1}{n(n+1)^2}$$

And since the first sum telescopes to $1$ and the identity $\frac{1}{n(n+1)^2}=\frac{1}{n(n+1)}-\frac{1}{(n+1)^2}$ holds we obtain that:

$$S=\sum_{n=1}^{\infty}\frac{1}{n(n+1)^2}=\sum_{n=1}^{\infty}\Big[\frac{1}{n(n+1)}-\frac{1}{(n+1)^2}\Big]=1-(\zeta(2)-1)=2-\frac{\pi^2}{6}$$

Answer 2

Continuing from where you left off, as suggested by Zarrax, $$ \begin{align} S&=\sum_{n=1}^\infty{1\over n(n+1)}\sum_{m=1}^n{1\over(m+1)^2}\\ &=\sum_{m=1}^\infty{1\over(m+1)^2}\sum_{n=m}^\infty{1\over n(n+1)}\\ &=\sum_{m=1}^\infty{1\over m(m+1)^2}\\ &=\sum_{m=1}^\infty\left({1\over m}-{1\over m+1}-{1\over(m+1)^2}\right)\\ &=\sum_{m=1}^\infty\left({1\over m}-{1\over m+1}\right)-\sum_{m=1}^\infty{1\over(m+1)^2}\\ &=1-\left({\pi^2\over6} -1\right)\\ &=2-{\pi^2\over6} \end{align} $$

Answer 3

A completely alternative approach can be *summed* up by writing the entire thing as a "single" sum:

$$S=\sum_{1\le k\le m\le n}f(k)f(m)f(n);\quad f(j)=\frac1{j(j+1)}$$

Notably, the summand is symmetric in $k,m,n$ i.e. swapping any two variables does not change the result. Using this, we can rewrite the summand with the bounds $(1\le m\le k\le n)$ or $(1\le n\le k\le m)$ etc. There are $3!=6$ ways to rearrange the bounds, but they all result in $S$. Adding every possible combination together thus gives us

$$6S=\sum_{k\ge1}\sum_{m\ge1}\sum_{n\ge1}f(k)f(m)f(n)+3\sum_{a\ge1}\sum_{b\ge1}f(a)f(b)^2+2\sum_{j\ge1}f(j)^3$$

The first triple series results from covering every possible $(k,m,n)$, since each $(k,m,n)$ must satisfy $(k\le m\le n)$ or some rearrangement of this. Then we have an extra double series for when two of the variables are equal, which has 6 possibilities, 3 of which are covered in the triple series. The last series comes about when all three variables are equal, which has 6 possibilities, 4 of which are given by the triple and double series.

Since all the indices are independent now, this is just a matter of doing some partial fractions and working each out. We have, from your work:

$$\sum_{n\ge1}f(n)=1$$

For the squared term, we have:

\begin{align}\sum_{n\ge1}f(n)^2&=\sum_{n\ge1}-\frac2n+\frac2{n+1}+\frac1{n^2}+\frac1{(n+1)^2}\\&=-\frac21-\frac1{1^2}+\sum_{n\ge1}\frac2{n^2}\\&=-3+\frac{\pi^2}3\end{align}

For the cubed term, we have:

\begin{align}\sum_{n\ge1}f(n)^3&=\sum_{n\ge1}\frac6n-\frac6{n+1}+\frac1{n^3}-\frac1{(n+1)^3}-\frac3{n^2}-\frac3{(n+1)^2}\\&=\frac61+\frac1{1^3}+\frac3{1^2}-\sum_{n\ge1}\frac6{n^2}\\&=10-\pi^2\end{align}

which finally results in

$$6S=1^3+3(1)\left(-3+\frac{\pi^2}3\right)+2(10-\pi^2)=12-\pi^2$$

$$S=2-\frac{\pi^2}6$$

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More explicit explanation

Let us define

$$S_1=\sum_{k,m,n\ge1}f(k)f(m)f(n)$$

$$S_2=\sum_{k,m\ge1}f(k)f(m)^2$$

$$S_3=\sum_{k\ge1}f(k)^3$$

It should be pretty clear that $f(k)f(m)f(n)$ should occur 6 times in our $6S$. We will tackle this case-by-case.

Consider the case of $k<m<n$. As an example, consider $f(1)f(2)f(3)$. How many times does this appear in $S_1$? Notice that it occurs for

$$(k,m,n)\in\{(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)\}$$

and every combination occurs in $S_1$.

Consider the case of $k\ne m=n$. As an example, consider $f(1)f(2)f(2)=f(1)f(2)^2$. How many times does this appear in $S_1$? Notice that it occurs for

$$(k,m,n)\in\{(1,2,2),(2,1,2),(2,2,1)\}$$

which is missing $3$ cases. The other $3$ case comes from $S_2$.

Consider the case of $k=m=n$. As an example, consider $f(1)f(1)f(1)=f(1)^3$. How many times does this appear in $S_1$? Notice it occurs only once, for $(k,m,n)=(1,1,1)$. How many times does this appear in $S_2$? It also occurs only once, for $(k,m)=(1,1)$. We multiply the $S_2$ by 3, which gives us 4 cases covered in total. We fill this in with 2 copies of $S_3$. This finally leaves us with

$$S=S_1+3S_2+2S_3$$

Answer 4

Just for your curiosity.

We could also have a cloased form for the partial sum $$S_p=\sum_{n=1}^{p}\sum_{m=1}^{n}\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}$$since $$\sum_{k=1}^{m}\frac{1}{(n+1)(k+1)(m+1)nmk}=\frac{1}{(m+1)^2 n (n+1)}$$ $$\sum_{m=1}^{n}\frac{1}{(m+1)^2 n (n+1)}=\frac{\pi ^2-6-6 \psi ^{(1)}(n+2)}{6 n (n+1)}$$ $$S_p=\sum_{n=1}^{p}\frac{\pi ^2-6-6 \psi ^{(1)}(n+2)}{6 n (n+1)}=\frac{12 (p+1)-\pi ^2 (p+2)+6 (p+2) \psi ^{(1)}(p+2)}{6 (p+1)}$$

Now, using, for large values of $q$, the expansion $$\psi ^{(1)}(q)=\frac{1}{q}+\frac{1}{2 q^2}+\frac{1}{6 q^3}+O\left(\frac{1}{q^5}\right)$$ and continuing with Taylor series for large values of $p$ $$S_p=\left(2-\frac{\pi ^2}{6}\right)+\frac{1-\frac{\pi ^2}{6}}{p}+\frac{\pi ^2-3}{6 p^2}-\frac{2+\pi ^2}{6 p^3}+\frac{10+\pi ^2}{6 p^4}+O\left(\frac{1}{p^5}\right)$$

For illustration purposes, using $p=10$, the exact value is $$S_{10}=\frac{42308191}{140873040}\approx 0.300328$$ while the above truncated expansion gives $$\frac{125690-10909 \pi ^2}{60000}\approx 0.300375$$

Answer 5

Yet another way: $$\sum_{1\leq k<m<n}\frac{1}{kmn(k+1)(m+1)(n+1)} = [x^3]\prod_{h\geq 1}\left(1+\frac{x}{h(h+1)}\right)=\frac{1}{\pi}[x^4]\cos\left(\frac{\pi}{2}\sqrt{1-4x}\right)$$ where the first equality follows from Viète's theorem and the second one from the Weierstrass product for the cosine function. The Maclaurin series of the RHS immediately leads to $$\sum_{1\leq k<m<n}\frac{1}{kmn(k+1)(m+1)(n+1)} = 5-\frac{\pi^2}{2} $$ and we may deal with the cases $k=m$ or $m=n$ in a similar fashion.

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\mbox{Evaluate}\ \bbox[15px,#ffd]{S \equiv \sum_{n = 1}^{\infty}\sum_{m = 1}^{n} \sum_{k = 1}^{m}{1 \over \pars{n + 1}\pars{k + 1}\pars{m + 1}nkm}}}$

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\begin{align} S & \equiv \sum_{n = 1}^{\infty}\sum_{m = 1}^{n} \sum_{k = 1}^{m}{1 \over \pars{n + 1}\pars{k + 1}\pars{m + 1}nkm} \\ & = \sum_{n = 1}^{\infty}{1 \over n\pars{n + 1}}\sum_{m = 1}^{n} {1 \over m\pars{m +1}}\ \overbrace{\sum_{k = 1}^{m}{1 \over k\pars{k + 1}}}^{\ds{m \over m + 1}} \\ &= \sum_{n = 1}^{\infty}{1 \over n\pars{n + 1}}\sum_{m = 1}^{n} {1 \over \pars{m + 1}^{2}} = \sum_{m = 1}^{\infty}{1 \over \pars{m + 1}^{2}}\ \overbrace{\sum_{n = m}^{\infty}{1 \over n\pars{n + 1}}}^{\ds{1 \over m}} \\[5mm] & = \sum_{m = 0}^{\infty}{1 \over \pars{m + 1}\pars{m + 2}^{2}} = \left.-\,\totald{}{\alpha} \sum_{m = 0}^{\infty}{1 \over \pars{m + 1}\pars{m + \alpha}} \right\vert_{\ \alpha\ =\ 2} \\[5mm] & = -\,\totald{}{\alpha}\bracks{\Psi\pars{\alpha} - \Psi\pars{1} \over \alpha - 1} _{\ \alpha\ =\ 2} \end{align}

$\ds{\Psi}$ is the Digamma Function and $\ds{\Psi\pars{1} = -\gamma}$ where $\ds{\gamma}$ is the Euler-Mascheroni Constant

Then, \begin{align} S & \equiv \sum_{n = 1}^{\infty}\sum_{m = 1}^{n} \sum_{k = 1}^{m}{1 \over \pars{n + 1}\pars{k + 1}\pars{m + 1}nkm} \\[5mm] & = \underbrace{\Psi\pars{2}}_{\ds{1 - \gamma}}\ +\ \gamma - \underbrace{\Psi\, '\pars{2}}_{\ds{{\pi^{2} \over 6} - 1}} = \bbox[15px,#ffd,border:1px solid navy]{2 - {\pi^{2} \over 6}}\ \approx 0.3551 \end{align}

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Note that $\ds{\Psi\pars{2}}$ is evaluated with its recursive property $\ds{\bf\color{black}{6.3.5}}$. Namely, $\ds{\Psi\pars{2} = \Psi\pars{\color{red}{1} + 1} = \Psi\pars{\color{red}{1}} + {1 \over \color{red}{1}} = -\gamma + 1}$.

In addition ( see $\ds{\ \bf\color{black}{6.3.16}}$ ): $\ds{\Psi\, '\pars{2} = \sum_{n = 1}^{\infty}{1 \over \pars{n + 1}^{2}} = \sum_{n = 2}^{\infty}{1 \over n^{2}} = \sum_{n = 1}^{\infty}{1 \over n^{2}} - {1 \over 1^{2}} = {\pi^{2} \over 6} - 1}$.