Nilakantha- Syed Double Series

The series can be simplified and can be written as $\sum _{ m=1 }^{ \infty }{ \sum _{ k=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ m } }{ { \left( 2m+1 \right) }^{ 2k+1 }\left( { \left( 2m+1 \right) }^{ 2 }-1 \right) } } } =\frac { 1 }{ 4 } \sum _{ m=1 }^{ \infty }{ \sum _{ k=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ m } }{ { \left( 2m+1 \right) }^{ 2k+1 }\left( m+1 \right) \left( m \right) } } }$ which can be further simplified as $\frac { 1 }{ 4 } \sum _{ m=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ m } }{ \left( 2m+1 \right) \left( m+1 \right) \left( m \right) } \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { \left( 2m+1 \right) }^{ 2k } } } }$ since second series is a geometric series so we get $\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { \left( 2m+1 \right) }^{ 2k } } } =\frac { 1 }{ { \left( 2m+1 \right) }^{ 2 }-1 } =\frac { 1 }{ 4\left( m+1 \right) \left( m \right) }$ on substitution we get $\frac { 1 }{ 16 } { \sum _{ m=1 }^{ \infty }{ \frac { { \left( -1 \right) }^{ m } }{ \left( 2m+1 \right) { \left( m+1 \right) }^{ 2 }{ \left( m \right) }^{ 2 } } } }$ by partial fraction decomposition and by evaluating each series we get $\frac { 1 }{ 16 } { \sum _{ m=1 }^{ \infty }{ \left( \frac { { 4\left( -1 \right) }^{ m+1 } }{ m } +\frac { { \left( -1 \right) }^{ m } }{ { m }^{ 2 } } +\frac { 4{ \left( -1 \right) }^{ m+1 } }{ m+1 } +\frac { { \left( -1 \right) }^{ m+1 } }{ { \left( m+1 \right) }^{ 2 } } +\frac { { 16\left( -1 \right) }^{ m } }{ \left( 2m+1 \right) } \right) } }=\frac { 1 }{ 16 } \left( (4\ln { 2)-\frac { { \pi }^{ 2 } }{ 12 } +4-4\ln { 2+1-\frac { { \pi }^{ 2 } }{ 12 } +4\pi -16 } } \right) =\frac { \pi }{ 4 } -\frac { { \pi }^{ 2 } }{ 96 } -\frac { 11 }{ 16 }$ therefore $\boxed{a+b+c+d+e+f=1+4+2+ 96+11+16 = 130}$

$\begin{aligned} S & = \sum_{m=0}^\infty \sum_{k=0}^\infty \frac {(-1)^{m+1}}{(2m+3)^{2k+3}\left((2m+3)^2-1\right)} \\ & = \sum_{m=0}^\infty \frac {(-1)^{m+1}}{(2m+3)^3\left((2m+3)^2-1\right)} \sum_{k=0}^\infty \left(\frac 1{(2m+3)^2}\right)^k \\ & = \sum_{m=0}^\infty \frac {(-1)^{m+1}}{(2m+3)^3\left((2m+3)^2-1\right)} \times \frac {(2m+3)^2}{(2m+3)^2 - 1} \\ & = \sum_{m=0}^\infty \frac {(-1)^{m+1}}{(2m+3)\left(4m^2+12m+8\right)^2} \\ & = \sum_{m=0}^\infty \frac {(-1)^{m+1}}{(2m+3)(2m+2)^2(2m+4)^2} & \small \blue{\text{By partial fraction decomposition}} \\ & = \sum_{m=0}^\infty \frac {(-1)^{m+1}}4 \left(\frac 1{(2m+2)^2} - \frac 2{2m+2} + \frac 4{2m+3} - \frac 1{(2m+4)^2} - \frac 2{2m+4} \right) \\ & = \sum_{m=0}^\infty \frac {(-1)^m}2 \left(\frac 1{2m+2} - \frac 2{2m+3} + \frac 1{2m+4} \right) - \sum_{m=0}^\infty \frac {(-1)^m}{16} \left(\frac 1{(m+1)^2} - \frac 1{(m+2)^2} \right) \\ & = \small \frac 12 \left(\frac 12 - \frac 23 + \frac 14 - \frac 14 + \frac 25 - \frac 16 + \frac 16 - \frac 27 + \cdots \right) - \frac 1{16} \left(\frac 1{1^2} - \frac 1{2^2} - \frac 1{2^2} + \frac 1{3^2} + \frac 1{3^2} - \frac 1{4^2} - \cdots \right) \\ & = \frac 14 - 1 + \sum_{n=0}^\infty \frac {(-1)^n}{2n+1} + \frac 1{16} - \frac 18 \sum_{n=1} \frac {(-1)^{n+1}}{n^2} & \small \blue{\text{Riemann zeta function }\zeta (s) = \sum_{k=1}^\infty \frac 1{k^s}} \\ & = - \frac {11}{16} + \tan^{-1} 1 - \frac 18 \left(\zeta (2) - \frac 12 \zeta(2)\right) & \small \blue{\text{and }\zeta (2) = \frac {\pi^2}6} \\ & = \frac \pi 4 - \frac {\pi^2}{96} - \frac {11}{16} \end{aligned}$

Therefore $a+b+c+d+e+f = 1 + 4 + 2 + 96 + 11 + 16 = \boxed{130}$ .

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Reference: Riemann zeta function