Summation splitting: $\sum_{n=1}^{\infty} \frac{n}{3\cdot 5\cdot 7 \dots (2n+1)}$

Solution 1:

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

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

Solution 2:

Let $x_n $ be the general term. Once you think the series telescopes, you want to write $x_n$ as $f(n) -f(n+1) $ for some function $f$. In particular, when you have a product in the denominator such as in this case, call it $\Pi(n)$ set $$x_n=\frac {An +B}{\Pi (n)}-\frac {A (n+1)+B}{\Pi(n+1)}.$$ In your case you'll get $A=0, B= 1/2$ that is $$ x_n=\frac12\left (\frac {1}{3\cdot\cdots(2n-1)}-\frac {1}{3\cdot\cdots(2n+1)}\right). $$If you then look at $S_N=\sum_{n=1}^N x_n $ you'll see that most of the terms cancel out leaving us $S_N=\frac12\left (1- \frac {1}{3\cdot\cdots(2N+1)}\right) $. Let $N\to\infty$ and you're done.