Calculate the value of the following series
$\begin{align} \sum_{k=1}^\infty \frac{1}{2\cdot(k-1)+1} \cdot \frac{1}{2\cdot(k)+1} \cdot \frac{1}{2\cdot(k+1)+1} \cdots \frac{1}{2\cdot(k+N)+1} \end{align}$
Where $N \in \mathbb{N}_0$
This should be similar to the telescoping series in this question.
And the answer should be $\begin{align} \frac{2^{N-1}}{N+1}\cdot\frac{N!}{(2N+1)!} \end{align}$
Any help is appreciated.
Well, if you already know how to proceed, just write it down.
\begin{align} \sum_{k=1}^{\infty}\frac1{(2k-1)\dots(2k+2N+1)} &= \frac1b\sum_{k=1}^{\infty}\frac{a+b-a}{(2k-1)\dots(2k+2N+1)} \\&= \frac1b\sum_{k=1}^{\infty}\frac{a+b}{(2k-1)\dots(2k+2N+1)} - \frac{a}{(2k-1)\dots(2k+2N+1)} \\&\stackrel{\text{?}}{=} \frac1b\sum_{k=1}^{\infty}\frac{1}{(2k-1)\dots(2k+2N-1)} - \frac{1}{(2k+1)\dots(2k+2N+1)} \end{align}
Expanding the last expression we have
$$\left\{ \begin{align} a+b &= 2k+2N+1\\ a &= 2k-1 \end{align} \right. $$
from which it follows that $b= 2(N+1)$. Notice that what matters in the last step is that $b$ does not depend on $k$.
Do you think you can take it from here?
