Using Raabe's test, examine the convergence of: sum
Using Raabe's test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n)!!} \cdot \frac{1}{2n+1}. $$
Explanation of Method
D'Alembert's ratio test on this series gives a limit of exactly $1$ (as for most double-factorial series), so Raabe's finer test is needed. Expressing $u_{n+1}$ in terms of $u_n$ via the double-factorial recursions $(2n+1)!!=(2n-1)!!(2n+1)$ and $(2n+2)!!=(2n)!!(2n+2)$ gives a clean rational expression for the ratio whose Raabe limit works out to $\frac32>1$.
- Express $u_{n+1}$ using the double-factorial recursions.
- Compute the ratio $u_n/u_{n+1}=\dfrac{(2n+2)(2n+3)}{(2n+1)^2}$.
- Evaluate the Raabe limit $r=\dfrac32>1$ and conclude convergence.
Step 1 · Set up $u_n$ and $u_{n+1}$
With $u_n=\dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac1{2n+1}$, using $(2n+1)!!=(2n-1)!!(2n+1)$ and $(2n+2)!!=(2n)!!(2n+2)$:
$$ u_{n+1} = \frac{(2n-1)!!(2n+1)}{(2n)!!(2n+2)}\cdot\frac1{2n+3}. $$
Step 2 · Compute the ratio
$$ \frac{u_n}{u_{n+1}} = \frac1{2n+1}\cdot\frac{2n+2}{2n+1}\cdot(2n+3) = \frac{(2n+2)(2n+3)}{(2n+1)^2}. $$
Step 3 · Apply Raabe's test
$$ \frac{u_n}{u_{n+1}}-1 = \frac{(2n+2)(2n+3)-(2n+1)^2}{(2n+1)^2} = \frac{6n+5}{(2n+1)^2}. $$
$$ r = \lim_{n\to\infty} n\cdot\frac{6n+5}{(2n+1)^2} = \lim_{n\to\infty}\frac{6n^2+5n}{4n^2+4n+1} = \frac64 = \frac32. $$
Since $r=\frac32>1$, by Raabe's test the series converges.
Summary Table
| Quantity | Value |
|---|---|
| $u_n/u_{n+1}$ | $\dfrac{(2n+2)(2n+3)}{(2n+1)^2}$ |
| Raabe limit $r$ | $3/2$ |
| Conclusion | Converges |
Additional Notes
- This series is the $x=1$ specialization of the classical Taylor expansion $\arcsin x=\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}\cdot\frac{x^{2n+1}}{2n+1}$, so its convergence at $x=1$ (the boundary of the radius of convergence) is exactly what this problem confirms directly, giving the identity $\frac\pi2=\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}\cdot\frac1{2n+1}$.
- Problem 92 examines the closely related series $\sum\left[\frac{(2n-1)!!}{(2n)!!}\right]^p$ and finds via Gauss's test that it converges only for $p>2$; this problem's extra factor of $\frac1{2n+1}$ effectively contributes an additional "$p=1$" worth of decay, nudging what would otherwise be a borderline case ($p=1$ alone diverges) into convergence.
- Since Wallis's formula gives $\frac{(2n-1)!!}{(2n)!!}\sim\frac1{\sqrt{\pi n}}$, the general term here behaves like $\frac1{2n\sqrt{\pi n}}\sim Cn^{-3/2}$, matching a convergent $p$-series with $p=3/2$, an independent confirmation of the Raabe result.
Result
The series converges: