Question

Find the value of $\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+.....$

Answer 1

$\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+......a_n=\frac{n}{1\cdot 3\cdot 5\cdot \mathrm{7.....}(2n+1)}{a}_n=\frac{1}{2}\left[\frac{2n+1-1}{1\cdot 3\cdot 5\cdot \mathrm{7.....}(2n+1)}\right]a_n=\frac{1}{2}[\frac{2n+1}{1\cdot 3\cdot 5\cdot \mathrm{7.....}(2n-1)(2n+1)}-\frac{1}{1\cdot 3\cdot 5\cdot \mathrm{7.....}(2n+1)}]a_n=\frac{1}{2}\{\frac{1}{1\cdot 3\cdot 5\cdot \mathrm{7.....}(2n-1)}-\frac{1}{1\cdot 3\cdot 5\cdot \mathrm{7.....}(2n+1)}\}{s}_n=t_1+t_2+t_3+t_4+.....t_n{s}_n=\frac{1}{2}\left[\right\{\frac{1}{1}-\frac{1}{1\cdot 3}\}+\{\frac{1}{1\cdot 3}-\frac{1}{1\cdot 3\cdot 5}\}+.......\{\frac{1}{1\cdot 3\cdot \mathrm{5.....}(2n-1)}-\frac{1}{1\cdot 3\cdot \mathrm{5.....}(2n+1)}\left\}\right]s_n=\frac{1}{2}[1-\frac{1}{1\cdot 3\cdot \mathrm{5.....}(2n-1)}]$

Now by definition , the sum $s$ of given infinite series is

$s={lim}_{n\to \infty }{s}_n{lim}_{n\to \infty }\frac{1}{2}[1-\frac{1}{1\cdot 3\cdot \mathrm{5.....}(2n-1)}]=\frac{1}{2}[1-0]=\frac{1}{2}$