Summation of general series

Question

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.

Answer 1

By computing the residues of $\frac{x^3}{x^4+4}$ in $x=\pm 1\pm i$ we have: $$\frac{x^3}{x^4+4}=\frac{1}{2}\left(\frac{x-1}{(x-1)^2+1}+\frac{x+1}{(x+1)^2+1}\right)\tag{1}$$ hence: $$\begin{eqnarray*}\sum_{n=0}^{k}\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}&=&\frac{1}{2}\sum_{n=0}^{k}(-1)^n\left(\frac{2n}{4n^2+1}+\frac{2(n+1)}{4(n+1)^2+1}\right)\\&=&\frac{(-1)^k\cdot(k+1)}{4(k+1)^2+1}.\end{eqnarray*}$$

Answer 2

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.