Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\left(\sqrt n+(-1)^{n-1}\right)^a}, \qquad a>0. $$
Explanation of Method
Expanding $(\sqrt n+(-1)^{n-1})^{-a}$ via the binomial-type asymptotic in the small parameter $(-1)^{n-1}/\sqrt n$ splits the term into a leading piece $n^{-a/2}$ times an alternating sign, plus a correction of relative order $n^{-1/2}$. This produces a genuine alternating $p$-series to leading order (absolute convergence for $a>2$, conditional for the rest via Leibniz after a Dirichlet-type regrouping) but shifted by the correction term, which requires $a>1$ on its own to converge absolutely — raising the true divergence boundary above what the naive leading-order analysis alone would suggest.
- Expand the term asymptotically in powers of $n^{-1/2}$.
- Classify the leading alternating term and the correction term separately.
- Combine to get the full three-way classification.
Step 1 · Asymptotic expansion
Write $\epsilon_n=(-1)^{n-1}$. Then $(\sqrt n+\epsilon_n)^{-a}=n^{-a/2}\left(1+\dfrac{\epsilon_n}{\sqrt n}\right)^{-a}=n^{-a/2}\left(1-\dfrac{a\epsilon_n}{\sqrt n}+O\!\left(\frac1n\right)\right)$. So:
$$ a_n = \epsilon_n\,n^{-a/2} - a\,n^{-(a+1)/2} + O\!\left(n^{-(a+2)/2}\right). $$
Step 2 · Classify each piece
The leading piece $\epsilon_n n^{-a/2}=(-1)^{n-1}/n^{a/2}$: absolutely convergent for $a/2>1$ i.e. $a>2$; conditionally convergent (Leibniz) for $0<a\le2$. The correction piece $-a\,n^{-(a+1)/2}$ has constant sign (no $\epsilon_n$ factor after the expansion — note this term does not alternate), so it converges (absolutely, being fixed-sign) iff $\frac{a+1}2>1$, i.e. $a>1$; for $a\le1$ it diverges to $-\infty$ term-by-term-summed. The remainder $O(n^{-(a+2)/2})$ converges absolutely whenever $a>-2$, always true here.
Step 3 · Combine
For $a>2$: leading term absolutely convergent, correction term absolutely convergent ($a>1$ holds) — overall absolutely convergent. For $1<a\le2$: leading term conditionally convergent, correction term absolutely convergent — overall conditionally convergent (since $|a_n|\sim n^{-a/2}$ whose series diverges for $a\le2$). For $0<a\le1$: correction term $-a\,n^{-(a+1)/2}$ diverges (fixed sign, $(a+1)/2\le1$), and since it is not oscillating it dominates and forces the whole series to diverge, despite the leading alternating term converging on its own.
Summary Table
| Range of $a$ | Conclusion |
|---|---|
| $a>2$ | Absolutely convergent |
| $1<a\le2$ | Conditionally convergent |
| $0<a\le1$ | Diverges |
Additional Notes
- The key subtlety, shared with Problem 134, is that a perturbation of an alternating term can introduce a non-alternating correction series whose own convergence threshold is stricter than the leading term's — here the correction $-a\,n^{-(a+1)/2}$ requires $a>1$, cutting off conditional convergence at $a=1$ rather than at $a=0$ as a naive Leibniz-only analysis of the leading term would suggest.
- This is structurally the archetype for a whole family of "perturbed alternating $p$-series" problems (compare Problem 134's logarithmic perturbation): whenever $(-1)^{n-1}$ is combined additively inside a power or transcendental function rather than as a pure multiplicative sign, a second, non-alternating series is generated by the expansion, and its threshold — not the naive leading term's — often determines the true lower boundary of convergence.
- At exactly $a=1$, both the leading term ($(-1)^{n-1}/\sqrt n$, conditionally convergent) and the correction term ($-1/n$, the harmonic series' divergence) are present, and the divergent harmonic-type correction wins — a clean illustration that "conditionally convergent plus divergent equals divergent," the same principle explored explicitly in Problem 136.
Result
Absolutely convergent for $a>2$; conditionally convergent for $1<a\le2$; diverges for $0<a\le1$: