Sum of $ \frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+ \cdots$ 
Finding series sum of $$ \frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\frac{1}{1\cdot 3\cdot 5\cdot 7}+\frac{1}{1\cdot 3\cdot 5\cdot 7\cdot 9}+ \cdots$$

Try: Let $\displaystyle a_{k}=\frac{1}{1\cdot 3\cdot 5\cdot 7\cdots (2k+1)}=\frac{2\cdot 4\cdot 6\cdots 2k}{(2k+1)!}$
So we have $\displaystyle a_{k}=\frac{2^k\cdot k!}{(2k+1)!}$
So our desired sum is $$\sum^{\infty}_{k=1}\frac{2^k\cdot k!}{(2k+1)!}$$
Now i am struck at that point 
I did not understand how can i solve further, 
Could some help me plaese , thanks

Let $$f(x)=\frac {x^3}{1.3}+\frac{x^5}{1.3.5}+...$$
We want the value of $f(1)$.  Note that 
$$\frac{df}{dx}=xf(x)+x^2,\\f(0)=0$$
