<s>[INST] 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 :) 
 [/INST] Your $a_n = \dfrac{n}{1 \cdot 3 \cdots (2n-1) \cdot (2n+1)} = \dfrac{n 2^n n!}{(2n+1)!}$.
Claim:
$$\sum_{n=1}^m \dfrac{n 2^n n!}{(2n+1)!} = \dfrac12 - \dfrac{2^{m-1} m!}{(2m+1)!}$$
The claim can be easily shown by induction.
Hence, the sum converges to $1/2$.
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