Summation of general series One of the problems in Donald Knuth's Art of Programming is phrased as follows:

Find and prove a simple formula for the sum $$\sum\limits_{n=0}^k\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}.$$

I have very little experience with summations. My method was to try to find some function such that
$$\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}=f(x)-f(x-1).$$
However, I have no idea how to do that systematically. Fortunately, partial fraction decomposition happens to give a function in that form, with
$$f(x)=\frac{(-1)^x(x+1)}{4(x+1)^2+1}$$
which is the solution in the back of the book. I was wondering how you might go about finding this solution. This specific problem is volume 1, chapter 1.2.1, problem 11.

Since
$$
x^4+4=(x^2+2-2x)(x^2+2+2x),
$$
we get
$$
f(x)=\frac{x^3}{x^4+4}=\frac{1}{4}\frac{(x^2+2-2x+x^2+2+2x)^2}{(x^2+2-2x)(x^2+2+2x)}.
$$
Let $a(x)=4x^2+1$. Then 
$$
4f(2n+1)=\frac{(a(n)+a(n+1))^3}{a(n)a(n+1)}=\frac{a(n)^2}{a(n+1)}+\frac{a(n+1)^2}{a(n)}+3(a(n)+a(n+1)).
$$
Hence
$$
\sum (-1)^nf(2n+1)=\sum (-1)^n\left\{ \frac{a(n)^2}{a(n+1)}+\frac{a(n+1)^2}{a(n)}+3(a(n)+a(n+1))\right\}
$$
As there is another answer appeared, I should stop here. Rearranging the sum, one can finish.
