Question

The sum to infinity of the series
$\frac{1}{3}+\frac{3}{3\cdot 7}+\frac{5}{3\cdot 7\cdot 11}+\frac{7}{3\cdot 7\cdot 11\cdot 15}+\cdots$ is

Answer 1

The $n^{\text{th}}$ term of the series is
$t_n=\frac{2n-1}{3\cdot 7\cdots (4n-1)}$
$t_n=\frac{1}{2}\left\{\frac{4n-1-1}{3\cdot 7\cdots (4n-1)}\right\}$
$t_n=\frac{1}{2}\{\frac{1}{3\cdot 7\cdots (4n-5)}-\frac{1}{3\cdot 7\cdots (4n-5)(4n-1)}\},n\ge 2$

Putting $n=2,3,...$ in succession
$t_2=\frac{1}{2}\{\frac{1}{3}-\frac{1}{3\cdot 7}\}$
$t_3=\frac{1}{2}\{\frac{1}{3\cdot 7}-\frac{1}{3\cdot 7\cdot 11}\}$
$⋮⋮$
$t_n=\frac{1}{2}\{\frac{1}{3\cdot 7\cdots (4n-5)}-\frac{1}{3\cdot 7\cdots (4n-5)(4n-1)}\}$

On adding, we get
$t_2+t_3+...+t_n=\frac{1}{2}\{\frac{1}{3}-\frac{1}{3\cdot 7\cdots (4n-1)}\}$

Sum to $n$ terms, $S=t_1+t_2+...+t_n$
$=\frac{1}{3}+\frac{1}{2}\{\frac{1}{3}-\frac{1}{3\cdot 7\cdots (4n-1)}\}$

$\underset{n\to \infty }{lim}S=\frac{1}{3}+\frac{1}{2}\cdot \frac{1}{3}=\frac{1}{2}$