How to find sum of $\frac{1}{2}+\frac{3}{8}+\frac{15}{48}+\frac{105}{384}+\cdots$?

Question

Sum $S_n$ of this series is given by, $S_n=\frac{1}{2}+\frac{3}{8}+\frac{15}{48}+\frac{105}{384}+...$

Express $r^{th}$ term ($U_r$) in this series and hence show that,$$\frac{U_r}{U_{r-1}}=\frac{2r-1}{2r}$$ Considering the above relation find $f_{(r)}$such that $U_{r-1}=f_{(r)}-f_{(r-1)}$. Here $f_{(r)}$shall be expressed interms of $U_r.$ Hence deduce $\sum_{r=1}^nU_r=(2n+1)U_n-1$

My Try:

I was able to find $U_r=\frac{(2r-1)!!}{(2r)!!}$

Simplifying further I got the next relation,

$U_r=\frac{(2r-1)(2r-3)!!}{2r(2r-2)!!}=\frac{(2r-1)}{(2r)}U_{r-1}$

But how do I proceed to find the summation. Please help me. Thanks in advance!

Answer 1

Hint:$$\frac{U_{r}}{U_{r-1}}=\frac{2r-1}{2r}$$

Rearrange $$U_{r-1}=\color{red}{2(r)U_{r}}-\color{blue}{2(r-1)U_{r-1}}=f_{r}-f_{r-1}$$

here $f_r=2rU_r$

Now telescope....