<s>[INST] Evaluate $\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $ $$\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $$
$$1 \cdot 3 \cdot 5 \cdots (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdots (2n+2)}{2 \cdot 4 \cdot 6 \cdots (2n+2)} = \frac{(2n+2)!}{2^{n+1} \cdot (n+1)!} $$
But the follow up gives me $\infty$. How to approach this type of exercises?
The result should be $1/2$
 [/INST] Observe that
$$
\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5\cdots (2n-1)}-\frac{1}{1\cdot3\cdot 5\cdots (2n+1)}\right)
$$
Hence
$$\sum_{n=1}^\infty\frac{n}{1\cdot 3\cdot 5\cdots (2n+1)}=\sum_{n=1}^\infty\frac{1}{2}\left(\frac{1}{1\cdot 3\cdot 5\cdots (2n-1)}-\frac{1}{1\cdot 3\cdot 5\cdots (2n+1)}\right)=\frac{1}{2}\left(1-\frac{1}{1\cdot3}+\frac{1}{1\cdot3}+\frac{1}{1\cdot3\cdot 5}-\frac{1}{1\cdot3\cdot 5}+\cdots\right)=\frac{1}{2}
$$
</s>