<s>[INST] 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?
 [/INST] 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 $1\cdot 3\cdot 5\cdot\ldots\cdot (2k+1)$.
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