Is there any better way of evaluating the summation

Question

Evaluate $$\sum_{r=1}^n\frac{r}{(2r-1)^2(2r+1)^2}$$

Solution that I have $$\sum_{r=1}^n\frac{r}{(2r-1)^2(2r+1)^2}=\sum_{r=1}^n\frac{r}{16r^4-8r^2+1}$$ $$=\color{red}{\sum_{r=1}^n\left(-\frac{1}{32\left(r+\frac12\right)^2}+\frac{1}{32\left(r-\frac12\right)^2}\right)}$$ Now this is a simple telescopic series yielding the answer $$\frac18-\frac{1}{32\left(n+\frac12\right)^2}=$$ $$\frac{n(n+1)}{2(2n+1)^2}$$ My question is how on earth one will imagine the thing written in coloured letters.

Is there any other nicer way$?$

Any help is greatly appreciated.

Answer 1

That is a partial fraction decomposition. $$ \frac{r}{(2r-1)^2(2r+1)^2} = \frac{1}{16}\frac{r}{(r-1/2)^2(r+1/2)^2} $$ where the denominator has two zeros with multiplicity two. The general theory of partial fraction decomposition states that $$ \frac{r}{(r-\frac 12)^2(r+\frac 12)^2} = \frac{A}{r-\frac 12} + \frac{B}{(r-\frac 12)^2} + \frac{C}{r+\frac 12} + \frac{D}{(r+\frac 12)^2} $$ with some constants $A, B, C, D$. These constants can be determined by multiplying with the common denominator and comparing coefficients. (There are also other method, see the above Wikipedia article.)

Here one gets $$ \frac{r}{(r-\frac 12)^2(r+\frac 12)^2} = \frac{1/2}{(r-\frac 12)^2} + \frac{-1/2}{(r+\frac 12)^2} \, . $$

Answer 2

Observe that

if $g(r)=2r+1, g(n-1)=2n-1, g(n)=? $

or if $f(r)=\dfrac1{\left(r+\dfrac12\right)^2}, f(n-1)=\dfrac1{\left(n-1+\dfrac12\right)^2}, f(n)=?$

So as $\left(r+\dfrac12\right)^2-\left(r+\dfrac12\right)^2=4\cdot r\cdot\dfrac12=?$

$$\implies\dfrac r{(2r-1)^2(2r+1)^2}=\dfrac{\left(r+\dfrac12\right)^2-\left(r+\dfrac12\right)^2}{32\left(r-\dfrac12\right)^2\cdot\left(r+\dfrac12\right)^2}=\cdots=\dfrac{f(r-1)-f(r)}{32}$$

$$\implies\sum_{r=1}^n\dfrac{f(r-1)-f(r)}{32}=\dfrac{f(0)-f(n)}{32}$$