Summing the sum

Calculus Level 5

$\large \sum _{ a=1 }^{ \infty }{ \sum _{ b=1 }^{ \infty }{ \dfrac { 1 }{ { a }^{ 2 }{ ( a+b ) }^{ 4 } } } } =-\dfrac { A }{ B } \zeta \left( C \right) +( { \zeta }( D ))^E$

The equation above holds true for positive integers $A,B,C,D$ and $E$ , where $A$ and $B$ are coprime. Find $A+B+C+D+E$ .

Notation : $\zeta(\cdot)$ denotes the Riemann zeta function .

Inspiration .

The answer is 18.

1 solution

Mark Hennings
Mar 3, 2016

The sum is $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^4} & = & \displaystyle \sum_{n=1}^\infty \left( \frac{H^{(2)}_{n+1}}{(n+1)^4} - \frac{1}{(n+1)^6}\right) \\ & = & \displaystyle \sum_{n=2}^\infty\left(\frac{H^{(2)}_n}{n^4} - \frac{1}{n^6}\right) \; = \; \sum_{n=1}^\infty\left(\frac{H^{(2)}_n}{n^4} - \frac{1}{n^6}\right) \\ & = & \displaystyle \zeta(3)^2 - \tfrac13\zeta(6) - \zeta(6) \; = \; -\tfrac43\zeta(6) + \zeta(3)^2 \end{array}$ making the answer $4 + 3 + 6 + 2 + 3 = \boxed{18}$ .

Typo in last line sir, it should be 18.

Nice solution(though if elaborate first step, it would be easier to catch.).

Harsh Shrivastava - 5 years, 3 months ago

Tell me how you convert into this form. Your very first step?

Ciara Sean - 5 years, 3 months ago

$\sum_{a=1}^\infty \sum_{b=1}^\infty \frac{1}{a^2(a+b)^4} \; =\; \sum_{a=1}^\infty \sum_{b=a+1}^\infty \frac{1}{a^2b^4} \; = \; \sum_{b=2}^\infty \sum_{a=1}^{b-1} \frac{1}{a^2b^4} \; = \; \sum_{b=2}^\infty \frac{H^{(2)}_{b-1}}{b^4} \; = \; \sum_{b=1}^\infty \frac{H^{(2)}_b}{(b+1)^4}$

Mark Hennings - 5 years, 3 months ago

thanks .....

Ciara Sean - 5 years, 2 months ago

Summing the sum

The answer is 18.

1 solution

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