Using Leibniz's test, prove the convergence of: sum
Using Leibniz's test, prove the convergence of: $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^\alpha}, \qquad \alpha>0. $$
Explanation of Method
Leibniz's test requires the positive sequence $b_n=1/n^\alpha$ to be decreasing with limit $0$ — both hold for every $\alpha>0$, since $x^{-\alpha}$ is a strictly decreasing function tending to $0$. So this alternating series converges unconditionally for all $\alpha>0$, even though the underlying $p$-series only converges absolutely when $\alpha>1$.
- Verify $b_n=1/n^\alpha$ is positive and decreasing for $\alpha>0$.
- Verify $b_n\to0$.
- Apply Leibniz's test and classify absolute vs. conditional convergence.
Step 1 · Positivity and monotonicity
$b_n=1/n^\alpha>0$ for all $n$. Since $f(x)=x^{-\alpha}$ has $f'(x)=-\alpha x^{-\alpha-1}<0$ for $\alpha>0$, $b_n$ is strictly decreasing.
Step 2 · Limit is zero
$$ \lim_{n\to\infty} \frac1{n^\alpha} = 0 \qquad \text{since } \alpha>0. $$
Step 3 · Apply Leibniz's test and classify
All three Leibniz conditions hold, so the series converges for every $\alpha>0$. Since $\sum|{(-1)^{n-1}}/n^\alpha|=\sum1/n^\alpha$ converges iff $\alpha>1$ (Problem 94), the series is absolutely convergent for $\alpha>1$ and only conditionally convergent for $0<\alpha\le1$.
Summary Table
| Range of $\alpha$ | Convergence type |
|---|---|
| $\alpha>1$ | Absolutely convergent |
| $0<\alpha\le1$ | Conditionally convergent |
Additional Notes
- The boundary case $\alpha=1$ is the classical alternating harmonic series, $\sum(-1)^{n-1}/n=\ln2$ — the most famous example of conditional convergence in all of analysis.
- This is the sharpest possible illustration of the gap between Leibniz's test and absolute convergence: the alternating series converges for the entire range $\alpha>0$, while the corresponding non-alternating $p$-series only converges for the strictly smaller range $\alpha>1$.
- Problem 111 revisits this exact series ($\alpha=1$) and shows that rearranging its terms changes the sum — a phenomenon (governed by the Riemann rearrangement theorem) that can only happen to conditionally convergent series, never to absolutely convergent ones.
Result
The series converges for every $\alpha>0$; absolutely if $\alpha>1$, conditionally if $0<\alpha\le1$: