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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

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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: $$\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!!}=\sqrt{\frac{\pi}{2}}\text{erf}\left(\frac{x}{\sqrt 2} \right) e^{\frac{x^2}{2}}$$ $$\Rightarrow S=\sqrt{\frac{e\pi}{2}}\text{erf}\left(\frac{1}{\sqrt 2}\right)$$ Where $\text{erf(z)}$ is the error function, defined as: $\displaystyle{\text{erf}(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}dt}$

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

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

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