Sum of $n$ terms and infinite terms of series 
The sum of $n$ terms of the series $$\frac{1}{2}+\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1\cdot 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1\cdot 3 \cdot 5}{4!}\left(\frac{1}{2}\right)^4+\frac{1\cdot 3 \cdot 5 \cdot 7}{5!}\left(\frac{1}{2}\right)^5+....$$
And also calculate sum of $\infty$ terms.

$\bf{My\; Try::}$ We can write above series as  $$ 1\underbrace{-\frac{1}{2}+\frac{1}{2!}\left(\frac{1}{2}\right)^2+\frac{1\cdot 3}{3!}\left(\frac{1}{2}\right)^3+\frac{1\cdot 3 \cdot 5}{4!}\left(\frac{1}{2}\right)^4+\frac{1\cdot 3 \cdot 5 \cdot 7}{5!}\left(\frac{1}{2}\right)^5+....}_{S_{n}}$$
So here $$\bf{r^{th}}\; terms \; of \; above\; series (T_{r}) = \frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-3)}{r!}\cdot \frac{1}{2^r}$$
So $$T_{r} = \frac{1}{3}\left[\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot  (2r-5)}{(r-1)!}\cdot \frac{1}{2^{r-1}}-\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-3)}{r!}\cdot \frac{1}{2^r}\right]$$
So $$S_{n} = \sum^{n}_{r=1}T_{r} = \frac{1}{3}\sum^{n}_{r=1}\left[\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot  (2r-5)}{(r-1)!}\cdot \frac{1}{2^{r-1}}-\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2r-3)}{r!}\cdot \frac{1}{2^r}\right]$$
So $$S_{n} = \frac{1}{3}\left[-3-\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2n-3)}{n!}\cdot \frac{1}{2^n}\right] = -1+\frac{1}{3}\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2n-3)}{n!}\cdot \frac{1}{2^n}$$
So our Sum is $$=\frac{1}{3}\frac{1\cdot 3 \cdot 5\cdot \cdot \cdot \cdot (2n-3)}{n!}\cdot \frac{1}{2^n}$$
Is my process is right or not, If not the how can i calculate it, Thanks
 A: 
This series looks like
  \begin{align*}
\frac{1}{2}&+\sum_{r=2}^\infty\frac{1}{r!}\left(\frac{1}{2}\right)^r(2r-3)!!\tag{1}\\
&=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}(2r-1)!!\\
&=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}\frac{(2r)!}{(2r)!!}\tag{2}\\
&=\frac{1}{2}+\sum_{r=1}^\infty\frac{1}{(r+1)!}\left(\frac{1}{2}\right)^{r+1}\frac{(2r)!}{2^rr!}\tag{3}\\
&=\frac{1}{2}+\frac{1}{2}\sum_{r=1}^\infty\frac{1}{r+1}\binom{2r}{r}\left(\frac{1}{4}\right)^r\tag{4}\\
&=\frac{1}{2}\sum_{r=0}^\infty\frac{1}{r+1}\binom{2r}{r}\left(\frac{1}{4}\right)^r\\
&=\frac{1}{2}\left.\left(\frac{1-\sqrt{1-4x}}{2x}\right)\right|_{x=\frac{1}{4}}\tag{5}\\
&=\frac{1}{2}\cdot 2\\
&=1
\end{align*}

Comment:


*

*In (1) we use double factorial notation $(2r-3)!!=(2r-3)\cdot(2r-5)\cdots 5\cdot 3\cdot 1$

*In (2) we use $r!=r!!\cdot(r-1)!!$

*In (3) we use $(2r)!!=2^r r!$

*In (4) we obtain the Catalan numbers $C_r=\frac{1}{r+1}\binom{2r}{r}$

*In (5) we use the generating function of the Catalan numbers
