Special summation 3 (edited)

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{n^4(n+1)^4} & \small \blue{\text{By partial fraction decomposition}} \\ & = \sum_{n=1}^\infty \left(\frac {20n^3+70n^2+84n+35}{(n+1)^4} - \frac {20n^3-10n^2+4n-1}{n^4}\right) \\ & = \sum_{n=1}^\infty \left(\frac {20}{n+1} + \frac {10}{(n+1)^2} + \frac 4{(n+1)^3} + \frac 1{(n+1)^4} - \frac {20}n + \frac {10}{n^2} - \frac 4{n^3} + \frac 1{n^4} \right) \\ & = \sum_{n=1}^\infty \left(\frac {20}{n+1} - \frac {20}n + \frac 4{(n+1)^3} - \frac 4{n^3}\right) + 2 \blue{\sum_{n=1}^\infty \left(\frac {10}{n^2} + \frac 1{n^4}\right)} - \frac {10}{1^2} - \frac 1{1^4} & \small \blue{\text{Riemann zeta function }\zeta(s) = \sum_{k=1}^\infty \frac 1{k^s}} \\ & = - 20 - 4 + 20 \blue{\zeta (2)} + 2 \blue{\zeta (4)} - 10 - 1 & \small \blue{\text{and }\zeta(2) = \frac {\pi^2}6, \zeta(4) = \frac {\pi^4}{90}} \\ & = \frac {\pi^4}{45} + \frac {10}3\pi^2 - 35 \end{aligned}$

Therefore, $\dfrac {EA}C + \dfrac BD = \dfrac {-35\times 3}{10\times 45} + \dfrac 42 = \dfrac {53}{30} \approx \boxed{1.767}$ .

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Reference:

• Partial fraction decomposition
• Riemann zeta function

$\quad\text{The sum} \\ = \displaystyle\sum_{n=1}^\infty \dfrac{1}{n^4(n+1)^4} \\ = \displaystyle\sum_{n=1}^\infty \left[\dfrac{1}{n(n+1)}\right]^4 \\ = \displaystyle\sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)^4 \\= \displaystyle\sum_{n=1}^\infty \left[\dfrac{1}{n^4}-\dfrac{4}{n^3(n+1)}+\dfrac{6}{n^2(n+1)^2}-\dfrac{4}{n(n+1)^3}+\dfrac{1}{(n+1)^4}\right] \\= \displaystyle\sum_{n=1}^\infty \left[\dfrac{1}{n^4}+\dfrac{1}{(n+1)^4}\right]- 4\displaystyle\sum_{n=1}^\infty \left[\dfrac{1}{n^3(n+1)}+\dfrac{1}{n(n+1)^3}\right] +6\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2(n+1)^2} \\= 2\zeta(4)-1-4\displaystyle\sum_{n=1}^\infty \dfrac{n^2+(n+1)^2}{n^3(n+1)^3}+6\left(\dfrac{\pi^2}{3}-3\right) \\= 2\times\dfrac{\pi^4}{90}-1-4\displaystyle\sum_{n=1}^\infty \dfrac{2n(n+1)+1}{n^3(n+1)^3}+2\pi^2-18 \\= \dfrac{\pi^4}{45}+2\pi^2-19-8\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2(n+1)^2}-4\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^3(n+1)^3} \\= \dfrac{\pi^4}{45}+2\pi^2-19-8\left(\dfrac{\pi^2}{3}-3\right)-4\left(10-\pi^2\right) \\= \dfrac{\pi^4}{45}+2\pi^2-19-\dfrac{8\pi^2}{3}+24-40+4\pi^2 \\= \dfrac{\pi^4}{45}+\dfrac{10}{3}\pi^2-35 \\ \therefore \dfrac{E}{C}\times A+\dfrac{B}{D} = -35\times\dfrac{3}{10}\times\dfrac{1}{45}+\dfrac{4}{2}=\dfrac{53}{30}\approx\boxed{1.767}$

I use the result of the question below:

Special summation 1

Special summation 2

I will do it with a little use of Einstein's notation to avoid summation signs. Say, \(\sum_{k=1}^\infty \frac{1}{(k(k+1))^4}

= \frac{1}{(k(k+1))^4}\)

So,

$\frac{1}{(k(k+1))^4}$

$=(\frac{1}{k}-\frac{1}{k+1})^4$

$=\frac{1}{k^4}+\frac{1}{(k+1)^4}-4(\frac{1}{k^3(k+1)}+\frac{1}{k(k+1)^3})+6(\frac{1}{(k(k+1))^2})$

$=\frac{1}{k^4}+\frac{1}{(k+1)^4}-4(\frac{1}{k^3}-\frac{1}{k^2(k+1)}+\frac{1}{k(k+1)^2}-\frac{1}{(k+1)^3})+6(\frac{1}{k^2}+\frac{1}{(k+1)^2}-2\frac{1}{k(k+1)})$

$=2\zeta(4)+12\zeta(2)-19-4(4-2\zeta(2))$

$=2\zeta(4)+20\zeta(2)-35=\frac{π^4}{45}+\frac{10π^2}{3}-35$ ).

[By the way $\zeta(2)=\frac{π²}{6} , \zeta(4)=\frac{π⁴}{90}$ ].

Hence, we have $A=\frac{1}{45},B=4,C=\frac{10}{3}, D=2, E=-35$ .

So, the given form has the value $2-\frac{7}{30}=1.7666666≈1.767$ .