Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty}(-1)^n\left[\frac{(2n-1)!!}{(2n)!!}\right]^a. $$
Explanation of Method
Using Wallis's asymptotic $\frac{(2n-1)!!}{(2n)!!}\sim\frac1{\sqrt{\pi n}}$ reduces the absolute series to a $p$-series with exponent $a/2$, giving the threshold $a>2$ for absolute convergence — matching Problem 92's Gauss's-test result exactly. Below that threshold, the underlying positive sequence is still monotonically decreasing to $0$ (a direct algebraic verification), so Leibniz's test rescues conditional convergence for the entire range $0<a\le2$.
- Use the Wallis asymptotic to identify the absolute series as $\sim\sum n^{-a/2}$.
- Determine the absolute-convergence threshold via the $p$-series test.
- Verify Leibniz's conditions for $0<a\le2$ to establish conditional convergence.
Step 1 · Asymptotic via Wallis's formula
Since $\dfrac{(2n-1)!!}{(2n)!!}\sim\dfrac1{\sqrt{\pi n}}$ (Problem 92), we get $b_n=\left[\dfrac{(2n-1)!!}{(2n)!!}\right]^a\sim\dfrac1{(\pi n)^{a/2}}$.
Step 2 · Absolute convergence threshold
$\sum b_n$ behaves like the $p$-series $\sum n^{-a/2}$, which converges iff $\frac a2>1$, i.e. $a>2$. So the series converges absolutely for $a>2$, and does not for $0<a\le2$.
Step 3 · Leibniz's test for $0<a\le2$
$b_n>0$ clearly. For monotonicity: $\dfrac{b_{n+1}}{b_n}=\left[\dfrac{2n+1}{2n+2}\right]^a<1$ for every $a>0$, so $b_n$ is decreasing. And $b_n\sim(\pi n)^{-a/2}\to0$ for $a>0$. By Leibniz's test, the series converges for every $a>0$; combined with Step 2, this convergence is conditional for $0<a\le2$.
Summary Table
| Range of $a$ | Conclusion |
|---|---|
| $a>2$ | Absolutely convergent |
| $0<a\le2$ | Conditionally convergent |
| $a\le0$ | Diverges ($b_n\not\to0$) |
Additional Notes
- This problem's threshold at $a=2$ exactly matches Problem 92's non-alternating series $\sum\left[\frac{(2n-1)!!}{(2n)!!}\right]^p$: the alternating sign here doesn't change the boundary at all, since the underlying decay rate (governed by Wallis's asymptotic) is what solely determines absolute convergence.
- The alternating sign does, however, dramatically extend the range of convergence: while the non-alternating Problem 92 series diverges for every $p\le2$, this alternating version converges (conditionally) throughout the entire range $0<a\le2$, illustrating the general power of Leibniz's test to rescue convergence wherever the underlying magnitude sequence is monotone decreasing to zero.
- The monotonicity proof here — $b_{n+1}/b_n=\left[\frac{2n+1}{2n+2}\right]^a<1$ for any $a>0$ — is a direct consequence of the double-factorial recursion, the same recursion used throughout Problems 86 and 92 for Raabe's and Gauss's tests.
Result
Absolutely convergent for $a>2$; conditionally convergent for $0<a\le2$: