about the following sum : $\frac{1}{1 \cdot 3}+\frac{2}{1 \cdot 3 \cdot5}++\ldots+\frac{n}{1 \cdot 3 \cdot 5 \cdot7 \cdot \ldots \cdot (2n+1)}$ Any hint about this expression : 
$$\frac{1}{1 \cdot 3}+\frac{2}{1 \cdot 3 \cdot5}+\frac{3}{1 \cdot3 \cdot5 \cdot7}+\ldots+\frac{n}{1 \cdot 3 \cdot 5 \cdot7 \cdot \ldots \cdot (2n+1)}$$
Thanks :) 
there must be a trick :) 

Note that
$$
\frac{2k}{1\cdot3\cdot5\cdots(2k+1)}
=\frac1{1\cdot3\cdot5\cdots(2k-1)}
-\frac1{1\cdot3\cdot5\cdots(2k+1)}\tag{1}
$$
Summing up both sides, where the right hand side telescopes, yields
$$
\sum_{k=1}^n\frac{2k}{1\cdot3\cdot5\cdots(2k+1)}
=1-\frac1{1\cdot3\cdot5\cdots(2n+1)}\tag{2}
$$
Letting $n\to\infty$ and dividing by $2$ yields
$$
\sum_{k=1}^\infty\frac{k}{1\cdot3\cdot5\cdots(2k+1)}=\frac12\tag{3}
$$
