# Convergence of Sum Involving Double Factorial

I have the sum $$\sum_{I=1}^\infty a^i (2i-1)!!$$

where $!!$ is the double factorial (the product of all the integers from 1 up to some non-negative integer n that have the same parity as n is called the double factorial).

I'd like to know how I should go about determining the values for $a$ that lead to convergence of the sum.

This is from a problem involving a ratio distribution, where I am trying to use the Taylor series to represent the inverse of the normal random variable in the denominator. Calculating the mean of the ratio distribution this way involves the sum of the second order moments of that r.v. I'd like to understand how to do the general form above and then apply it to this application to determine what conditions need to exist for convergence.

• If $a \ne 0$, the general term doesn't tend to zero, so the series diverges. The quotient of two consecutive terms of index $i, i+1$ is $a(2i-1)$. – David Jan 8 '16 at 22:05

Applying the ratio test to the terms $a_i$, we have

\begin{align*} \frac{a_{i + 1}}{a_i} &= \frac{a^{i + 1} (2i + 1)!!}{a^i (2i - 1)!!} = a (2i + 1) \end{align*}

This blows up in absolute value unless $a = 0$. (As a consequence of this estimate, the terms of the series don't even tend to zero)

The moral of the story is that things that are at all like a factorial are really big.

• Yes, this looks pretty straightforward. I'll have to find some other explanation why my Monte Carlo simulations seem to converge. I may have to post a different question about my rationale that took me to this expression. – Jim Jan 8 '16 at 22:26

This sum will only converge for $a=0$.

Using the ratio test we can determine the radius of convergence. Consider

$$L=\lim_{n\to \infty} \frac{|a|^{n+1} (2(n+1)-1)!!}{|a|^n (2n-1)!!}.$$

Now since both $(2(n+1)-1)!!$ and $(2n-1)!!$ have the same parity, the quotient simplifies nicely to

$$L=\lim_{n\to \infty} |a| (2n+1) = \infty.$$

Thus the only way this limit can be less than 1 (and thus resulting in a convergent series) is if $a=0$.

Moreover, the radius of convergence is given by $R=1/L=0$.

Thus the series only converges at zero.