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

Try: Let $displaystyle a_{k}=frac{1}{1cdot 3cdot 5cdot 7cdots (2k+1)}=frac{2cdot 4cdot 6cdots 2k}{(2k+1)!}$

So we have $displaystyle a_{k}=frac{2^kcdot k!}{(2k+1)!}$

So our desired sum is $$sum^{infty}_{k=1}frac{2^kcdot k!}{(2k+1)!}$$

Now i am struck at that point

I did not understand how can i solve further,

Could some help me plaese , thanks

That term in the denominator is also called the double factorial, namely: $$(2n+1)!!=frac{(2n+1)!}{2^n n!}$$
And our series can be rewritten as: $$S=sum_{n=0}^infty frac{1}{(2n+1)!!}$$
Now from this link see (21), is given that: $$sum_{n=0}^infty frac{x^{2n+1}}{(2n+1)!!}=sqrt{frac{pi}{2}}text{erf}left(frac{x}{sqrt 2} right) e^{frac{x^2}{2}}$$
$$Rightarrow S=sqrt{frac{epi}{2}}text{erf}left(frac{1}{sqrt 2}right)$$ Where $text{erf(z)}$ is the error function, defined as: $displaystyle{text{erf}(z)=frac{2}{sqrt{pi}}int_0^z e^{-t^2}dt}$

Let $$f(x)=frac {x^3}{1.3}+frac{x^5}{1.3.5}+...$$
We want the value of $f(1)$. Note that

$$frac{df}{dx}=xf(x)+x^2,\f(0)=0$$