the summation of $-1^k(1\cdot3\cdot5\cdots(2k-1))x^{2k}$

Question

I want to figure how and according to which formula is the summation

$$\sum_{k=0}(-1)^k(1\cdot3\cdot5\cdots(2k-1))x^{2k}$$

according to which formula this equality is true

Answer 1

Assume $x\ne0$.

Then by applying the ratio test to this power series, one has $$ \begin{align} L = \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|&=\lim_{k\to\infty}\left|\frac{(-1)^{k+1}(1\cdot3\cdot5\cdots(2k-1)(2k+1))}{(-1)^k(1\cdot3\cdot5\cdots(2k-1))}\cdot x^2\right| \\\\&=\lim_{k\to\infty}(2k+1)|x|^2 \\\\&=\infty, \end{align} $$ the radius of convergence is then $R=0$.

The given series never converges except for $x=0$.

Answer 2

Altho Olivier has already given you a good answer, you might nevertheless be interested in the following asymptotic expansion:

$$\sqrt{\frac{\pi z}{2}}\exp\left(\frac{z}{2}\right)\operatorname{erfc}\left(\sqrt{\frac{z}{2}}\right)\sim\sum_{k=0}^\infty\frac{(-1)^k (2k-1)!!}{z^k}$$

where $k!!$ is a double factorial, and $\operatorname{erfc}$ is the complementary error function. This result can be obtained in a manner similar to this answer, where the appropriate Mellin transform identity is

$$\int_0^\infty u^k\frac{\exp\left(-\frac{u}{2}\right)}{\sqrt{u}} \, \mathrm du=2^{k+\frac12}\Gamma\left(k+\frac12\right)=\sqrt{2\pi}(2k-1)!!$$