Nilakantha Series

$\begin{aligned} S & = \sum_{n=0}^\infty \frac {(-1)^n}{(2n+3)^3 - (2n+3)} \\ & = \sum_{n=0}^\infty \frac {(-1)^n}{(2n+3)\left(4n^2+12n+8\right)} \\ & = \sum_{n=0}^\infty \frac {(-1)^n}{(2n+3)(2n+2)(2n+4)} & \small \blue{\text{By partial fraction decomposition}} \\ & = \sum_{n=0}^\infty \frac {(-1)^n}2 \left(\frac 1{2n+2} - \frac 2{2n+3} + \frac 1{2n+4} \right) \\ & = \frac 12 \left(\frac 12 - \frac 23 + \frac 14 - \frac 14 + \frac 25 - \frac 16 + \frac 16 - \frac 27 + \frac 18 - \cdots \right) \\ & = \frac 14 - 1 + \left(1 - \frac 13 + \frac 15 - \frac 17 + \cdots \right) \\ & = \frac 14 - 1 + \tan^{-1} 1 \\ & = \frac {\pi - 3}4 \end{aligned}$

Therefore $a+b+c = 1+3+4 = \boxed 8$ .

The series can be written in the form $\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n } }{ \left( 2n+3 \right) \left( { \left( 2n+3 \right) }^{ 2 }-1 \right) } } =\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n } }{ \left( 2n+3 \right) \left( 2n+2 \right) \left( 2n+4 \right) } }$ by partial fraction decomposition we can further simplify it as $\frac { 1 }{ 4 } \sum _{ n=0 }^{ \infty }{ \left( \frac {{ \left( -1 \right) }^{ n } }{ (n+1) } +\frac { { \left( -1 \right) }^{ n } }{( n+2) } +\frac { { 4\left( -1 \right) }^{ n+1 } }{ 2n+3 } \right) }$ where $\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n } }{ n+1 } } =\ln { \left( 2 \right) } ;\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n } }{ n+2 } =1-\ln { 2;\sum _{ n=0 }^{ \infty }{ \frac { 4{ \left( -1 \right) }^{ n+1 } }{ \left( 2n+3 \right) } =\pi } } } -4$ on substituting we get $\frac { 1 }{ 4 } \left( \ln { 2+1-\ln { 2+\pi -4 } } \right) =\frac { \pi -3 }{ 4 }$ therefore we get $a=1$ , $b=3$ , $c=4$ so $\boxed{a+b+c=8}$