I encounter this series in an old book of infinite series. The series is as followed:

$1+(\dfrac{1}{2})x^2+(\dfrac{1\cdot 3}{2\cdot4})^2x^4+(\dfrac{1\cdot3\cdot5}{2\cdot4\cdot6})^3x^6+(\dfrac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8})^4x^8...$

Judging from the series, it must be an even function. Is it possible to find the closed form function for this series. Also, instead of the power of 2 in coefficients, we may have them in n. Do we have a general formula to generate the functions for power of n?

I try to differentiate this series but it doesn't seem to produce anything fruitful.

  • $\begingroup$ The exponents on the coefficients are really annoying. If I am right, by Stirling, the coefficients (with exponents) are asymptotic to $n^{-n/2}$. $\endgroup$
    – user65203
    Nov 6, 2019 at 17:49
  • $\begingroup$ @Yves Daoust Yeah, you are right, without the exponent things may become much easier. $\endgroup$ Nov 6, 2019 at 17:49

1 Answer 1


Looks like

$$\sum_{n=0}^\infty \bigg(\frac{(2n+1)!!}{(2n)!!}x^2\bigg)^n.$$

  • $\begingroup$ Chin: this is for the general series that I ask right? $\endgroup$ Nov 6, 2019 at 17:23
  • $\begingroup$ This is for the exact series that you have provided. $\endgroup$ Nov 6, 2019 at 17:23
  • $\begingroup$ Do you know how to find the closed form function for this series? $\endgroup$ Nov 6, 2019 at 17:25
  • $\begingroup$ I cannot imagine there should be one. WolframAlpha is unable to provide me with anything. $\endgroup$ Nov 6, 2019 at 17:28
  • $\begingroup$ Not all power series have closed form expression, right? $\endgroup$ Nov 6, 2019 at 17:29

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