I have got the following series $$\sum_{k=0}^\infty \frac{1}{(2n-1)(2n+1)(2n+3)}$$
I'm trying to expand it as a telescoping series and then calculate the partial sum series but didn't succeed so far. If someone can help me expand it as a telescoping series and explain what's the technique for it it would be great.
Thanks.
$$\dfrac{2n+3-(2n-1)}{(2n+3)(2n+1)(2n-1)}$$
$$=\dfrac{1}{(2n+1)(2n-1)}-\dfrac{1}{(2n+3)(2n+1)}$$
$$=F_n-F_{n+1}$$
where $$F_m=\dfrac{1}{(2m+1)(2m-1)}$$
HINT:
Using partial fraction expansion, we can write
$$\begin{align} \frac{1}{(2n-1)(2n+1)(2n+3)}&=\frac18\left(\frac{1}{2n-1}-\frac{2}{2n+1}+\frac{1}{2n+3}\right)\\\\ &=\frac18 \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)+\frac18 \left(\frac{1}{2n+3}-\frac{1}{2n+1}\right) \end{align}$$
Another effective approach exploits geometric series and Euler's beta function: $$\begin{eqnarray*} \sum_{k\geq 0}\frac{1}{(2k-1)(2k+1)(2k+3)}&=&\frac{1}{8}\sum_{k\geq 0}\frac{1}{\left(k-\frac{1}{2}\right)\left(k+\frac{1}{2}\right)\left(k+\frac{3}{2}\right)}\\&=&\frac{1}{8}\sum_{k\geq 0}\frac{\Gamma\left(k-\frac{1}{2}\right)}{\Gamma\left(k+\frac{5}{2}\right)}\\&=&\frac{1}{8\,\Gamma(3)}\sum_{k\geq 0}B\left(3,k-\frac{1}{2}\right)\\&=&\frac{1}{16}\int_{0}^{1}\sum_{k\geq 0}(1-x)^2 x^{k-3/2}\,dx\\&=&\frac{1}{16}\int_{0}^{1}(1-x) x^{-3/2}\,dx\\&=&\frac{B(2,-1/2)}{16}=\color{red}{-\frac{1}{4}.}\end{eqnarray*}$$
The technique is called partial fractions:
$$\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{A}{(2n-1)}+\frac{B}{2n+1}+\frac{C}{2n+3}$$
Now combine the denominators and equate coefficients.
$$1=A(2n+3)(2n+1)+B(2n-1)(2n+3)+C(2n-1)(2n+1)$$
$$1=A(4n^2+8n+3)+B(4n^2+4n-3)+C(4n^2-1)$$
We get:
$$4A+4B+4C=0$$
$$8A+4B=0$$
$$3A-3B-C=1$$
Substitution is often used to solve these types of equations. Note dividing the first equation leaves us with:
$$A+B+C=0$$
Hence:
$$A+B=-C$$
Substituting for $-C$ in the third equation gives:
$$3A-3B+A+B=1$$
$$4A-2B=1$$
But from the second equation we now have:
$$8A+4B=0$$
$$4A-2B=1$$
Multiply the $4A-2B=1$ equation by $2$ on both sides:
$$8A+4B=0$$
$$8A-4B=2$$
Add both equations above,
$$16A=2$$
So
$$A=\frac{1}{8}$$
Substituting this back into one of the equations that we just used elimination on we get:
$$B=-\frac{1}{4}$$
Now remember:
$$A+B=-C$$
So
$$-(A+B)=C$$
And hence,
$$C=\frac{1}{8}$$
Therefore,
$$\frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{(1/8)}{(2n-1)}-\frac{2/8}{2n+1}+\frac{1/8}{2n+3}=\frac{1/8}{2n-1}-\frac{1/8}{2n+1}+\frac{1/8}{2n+3}-\frac{1/8}{2n+1}$$
I broke it down even more so it would be easier to see the cancellations.
