Sum of n terms of the series $\frac{1}{1 \cdot 3}+\frac{2}{1 \cdot 3 \cdot5}+\frac{3}{1 \cdot 3 \cdot 5 \cdot 7}+\cdots$ I need to find the sum of n terms of the series 
$$\frac{1}{1\cdot3}+\frac{2}{1\cdot 3\cdot 5}+\frac{3}{1\cdot 3\cdot 5\cdot 7}+\cdots$$
And I've no idea how to move on. It doesn't look like an arithmetic progression  or a geometric progression. As far as I can tell it's not telescoping. What do I do?
 A: Clearly $$U_{r+1}=\frac{r}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
$$2U_{r+1}=\frac{2r}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
$$2U_{r+1}=\frac{(2r+1)-1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
$$2U_{r+1}=\frac{(2r+1)}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}-\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}$$
$$2U_{r+1}=\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}-\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
Now let $$V_r=\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)}$$
Then $$V_{r+1}=\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
Thus $$2U_{r+1}=V_r-V_{r+1}$$
$$\displaystyle 2\sum_{r=1}^{n} U_{r+1}=\sum_{r=1}^{n} \{V_r-V_{r+1}\}=V_1-V_{n+1}$$
$$\displaystyle 2\sum_{r=1}^{n} U_{r+1}=V_1-V_{n+1}$$
$$\displaystyle 2\sum_{r=1}^{n} U_{r+1}=\frac{1}{1}-\frac{1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2n-3)\cdot(2n-1)\cdot(2n+1)}$$
A: It is telescoping. Consider that:
$$ \frac{1}{1\cdot 3} = \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right),\quad \frac{2}{1\cdot 3\cdot 5} = \frac{1}{2}\left(\frac{1}{1\cdot 3}-\frac{1}{3\cdot 5}\right), $$
$$ \frac{3}{1\cdot 3\cdot 5\cdot 7}=\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5}-\frac{1}{3\cdot 5\cdot 7}\right),\quad \ldots$$
so:
$$ \sum_{k=1}^{n}\frac{k}{(2k+1)!!} = \frac{1}{2}\left(1-\frac{1}{(2n+1)!!}\right). $$
As usual, $(2k+1)!!$ stands for $1\cdot 3\cdot 5\cdot\ldots\cdot (2k+1)$.
