Using Gauss's test, examine the convergence of: sum
Using Gauss's test, examine the convergence of: $$ \sum_{n=1}^{\infty}\left[\frac{(2n-1)!!}{(2n)!!}\right]^p. $$
Explanation of Method
The plain ratio $u_n/u_{n+1}=\left(\frac{2n+2}{2n+1}\right)^p$ has Raabe limit $h=p/2$, which is inconclusive precisely at $p=2$ ($h=1$). Gauss's test resolves this exactly by expanding one order further, extracting the $1/n^2$ coefficient $A=\frac{p(p-1)}8$, which at $p=2$ equals $\frac14<1$ — settling the boundary case as divergent and confirming the clean threshold $p>2$.
- Compute the ratio $u_n/u_{n+1}=\left(\frac{2n+2}{2n+1}\right)^p$.
- Expand via the binomial series to get $h=p/2$ and $A=p(p-1)/8$.
- Apply Gauss's test, including the boundary case $p=2$.
Step 1 · Compute the ratio
Using $(2n+1)!!=(2n-1)!!(2n+1)$ and $(2n+2)!!=(2n)!!(2n+2)$, $u_{n+1}=u_n\left[\frac{2n+1}{2n+2}\right]^p$, so:
$$ \frac{u_n}{u_{n+1}} = \left[\frac{2n+2}{2n+1}\right]^p = \left(1+\frac1{2n+1}\right)^p. $$
Step 2 · Expand to second order
With $\frac1{2n+1}=\frac1{2n}-\frac1{4n^2}+O(1/n^3)$ and the binomial expansion $(1+x)^p=1+px+\frac{p(p-1)}2x^2+\cdots$:
$$ \frac{u_n}{u_{n+1}} = 1 + \frac{p/2}{n} + \frac{p(p-1)/8}{n^2} + O(1/n^3). $$
Step 3 · Apply Gauss's test
Here $h=\frac p2$ and $A=\frac{p(p-1)}8$. If $p>2$: $h>1$, converges. If $p<2$: $h<1$, diverges. If $p=2$: $h=1$, and $A=\frac{2\cdot1}8=\frac14<1$, so the series diverges.
Summary Table
| Value of $p$ | Conclusion |
|---|---|
| $p>2$ | Converges ($h=p/2>1$) |
| $p=2$ | Diverges ($h=1$, $A=1/4<1$) |
| $p<2$ | Diverges ($h=p/2<1$) |
Additional Notes
- By Wallis's product, $\frac{(2n-1)!!}{(2n)!!}\sim\frac1{\sqrt{\pi n}}$, so $u_n\sim\frac1{(\pi n)^{p/2}}$: this is precisely a $p$-series with exponent $p/2$, which converges iff $p/2>1$, i.e. $p>2$ — an independent confirmation matching Gauss's test exactly, including the boundary.
- This problem generalizes Problem 86 (the case "$p=1$ times an extra $\frac1{2n+1}$ factor"): here with no extra factor, the plain power $p$ alone must exceed the higher threshold $2$, rather than $1$, since removing the $\frac1{2n+1}$ decay shifts the whole convergence boundary.
- The boundary case $p=2$ is a clean illustration of why Gauss's test is strictly more powerful than Raabe's alone: Raabe's test only says $h=1$ is inconclusive, but the second-order coefficient $A=1/4$ immediately resolves it without needing to fall back on the Wallis-product asymptotic.
Result
The series converges if and only if $p>2$: