# value of $\sum^{\infty}_{k=0}\frac{(-1)^k}{\bigg(1\cdot 3 \cdot 5\cdots \cdots (2k+1)\bigg)\cdot 2^k}$

Finding value of $$\displaystyle \sum^{\infty}_{k=0}\frac{(-1)^k}{\bigg(1\cdot 3 \cdot 5\cdots \cdots (2k+1)\bigg)\cdot 2^k}$$

Try: We can write it as $$\sum^{\infty}_{k=0}\frac{(-1)^k\cdot\bigg( 2\cdot 4 \cdot 6\cdots (2k)\bigg)}{\bigg(1\cdot 2 \cdot 3\cdots (2k+1)\bigg)\cdot 2^k}=\sum^{\infty}_{k=0}\frac{(-1)^k\cdot 2^k\cdot k!}{(2k+1)!\cdot 2^k}$$

So we have $$\sum^{\infty}_{k=0}\frac{(-1)^k\cdot k!\cdot (k+1)!}{(2k+1)!\times (k+1)!}=\sum^{\infty}_{k=0}\frac{(-1)^k}{\binom{2k+1}{k}\cdot (k+1)!}$$

Now i did not understand how to solve further

Could some help me , Thanks

## 1 Answer

Rewriting the sum as $$2\sum_{k=0}^\infty\frac{(-1)^k2^k}{(2k+1)!!}\left(\frac12\right)^{2k+1}$$ we see the Maclaurin series of the Dawson integral, with argument $$\frac12$$: $$=2F\left(\frac12\right)=\sqrt\pi e^{-\frac14}\operatorname{erfi}\left(\frac12\right)$$