Summation splitting: $\sum_{n=1}^{\infty} \frac{n}{3\cdot 5\cdot 7 \dots (2n+1)}$ $$\sum_{n=1}^{\infty} \frac{n}{3\cdot 5\cdot 7 \dots (2n+1)}$$
Can someone please help me in solving this problem, I tried to take the summation of the numerator and denominator individually but my teacher said that it is wrong to do the summation individually, can somebody please explain why is it wrong to take individual summation and please recommend the correct way of solving this problem

Hint. One may observe that, for $n\ge1$,
$$
\begin{align}
\frac {n}{1\cdot 3\cdots (2n+1)} &=\color{red}{\frac12} \cdot \frac {(2n+1)-1}{1\cdot 3\cdots (2n+1)} 
\\\\&=\color{red}{\frac12} \cdot \frac {1}{1\cdot 3\cdots (2n-1)}-\color{red}{\frac12} \cdot \frac {1}{1\cdot 3\cdots (2n+1)},
\end{align}
$$ then, by telescoping terms, one obtains

$$
\sum_{n=1}^\infty\frac {n}{1\cdot 3\cdots (2n+1)}=\color{red}{\frac12}.
$$

