Find a formula for $\sum\limits_{n=0}^N(-1)^n\frac{(2n+1)^3}{(2n+1)^4+4}$ I found this sum in the mathematial induction chapter of The art of Computer Programming and i have no idea how to solve it.
$\dfrac{1^3}{1^4+4}-\dfrac{3^3}{3^4+4} +  ... +\dfrac{(-1)^n(2n+1)^3}{(2n+1)^4+4} $
I tried writing it as $\dfrac{1^3}{1*1^3+4}-\dfrac{3^3}{3*3^3+4} +  ... +\dfrac{(-1)^n(2n+1)^3}{(2n+1)*(2n+1)^3+4} $
and then writing it as
$\dfrac{1}{1*1^3+4}-\dfrac{3+5}{3*3^3+4} +...+\dfrac{(-1)^n(((2n+1)^2-(2n+1)+1)+...+((2n+1)^2+(2n+1)-1))}{(2n+1)*(2n+1)^3+4} $
but did not know how to continue.
I also tried writing it as
$\dfrac{1}{1+\dfrac{4}{1^3}}-\dfrac{1}{3+\dfrac{4}{3^3}} +  ... +\dfrac{(-1)^n}{2n+1+\dfrac{4}{(2n+1)^3}} $ but without succes.


Keyword: Concatenation.

First note that, for every $x$, $$\frac{4x^3}{x^4+4}=\frac{4x^3}{(x^2+2)^2-4x^2}=\frac{x^2}{(x-1)^2+1}-\frac{x^2}{(x+1)^2+1}$$ hence the partial sum $S_{2N+1}$ of $2N+1$ terms is such that $$4S_{2N+1}=\frac45+\sum_{n=1}^N\left(\frac{4(4n+1)^3}{(4n+1)^4+4}-\frac{4(4n-1)^3}{(4n-1)^4+4}\right)$$ that is, $$4S_{2N+1}=\frac45+U_N-V_N$$
where $$U_N=\sum_{n=1}^N\frac{(4n+1)^2}{(4n)^2+1}+\frac{(4n-1)^2}{(4n)^2+1}=\sum_{n=1}^N\frac{2((4n)^2+1)}{(4n)^2+1}=2N$$ and $$V_N=\sum_{n=1}^N\frac{(4n+1)^2}{(4n+2)^2+1}+\frac{(4n-1)^2}{(4n-2)^2+1}$$ Thus, $$V_N=\frac95-\frac{(4N+3)^2}{(4N+2)^2+1}+\sum_{n=1}^N\frac{(4n+1)^2}{(4n+2)^2+1}+\frac{(4n+3)^2}{(4n+2)^2+1}$$ that is, $$V_N=\frac95-\frac{(4N+3)^2}{(4N+2)^2+1}+\sum_{n=1}^N\frac{2((4n+2)^2+1)}{(4n+2)^2+1}=\frac95-\frac{(4N+3)^2}{(4N+2)^2+1}+2N$$ Coming back to $S_{2N+1}$, one gets $$4S_{2N+1}=\frac45-\frac95+\frac{(4N+3)^2}{(4N+2)^2+1}=\frac{8N+4}{(4N+2)^2+1}$$ hence, finally, 

$$S_{2N+1}=\frac{2N+1}{4(2N+1)^2+1}$$

The even numbered sums $S_{2N}$ are solvable by a similar treatment.
Fun fact: $$\lim_{N\to\infty}S_N=0$$
