Using Abel's test, prove the convergence of: sum from n=2
Using Abel's test, prove the convergence of: $$ \sum_{n=2}^{\infty}(-1)^n \frac{\sqrt[n]{n}}{\ln n}. $$
Explanation of Method
Factoring as $a_n=\frac{(-1)^n}{\ln n}$ (convergent by Leibniz's test, as in Problems 108 and 119) and $b_n=n^{1/n}$ sets up Abel's test. Since $n^{1/n}$ decreases to $1$ once $n>e$ (the same threshold that appears in Problem 70's root-test limit computation), it is eventually monotone and bounded between $1$ and its maximum value $\sqrt[3]3$, satisfying Abel's requirements.
- Show $\sum(-1)^n/\ln n$ converges via Leibniz's test.
- Show $b_n=n^{1/n}$ is eventually decreasing (for $n\ge3$).
- Show $b_n$ is bounded, and apply Abel's test.
Step 1 · Convergence of $\sum(-1)^n/\ln n$
As in Problem 108: $\frac1{\ln n}>0$, decreasing, and $\to0$. By Leibniz's test, $\sum_{n=2}^{\infty}\dfrac{(-1)^n}{\ln n}$ converges.
Step 2 · Eventual monotonicity of $b_n=n^{1/n}$
With $f(x)=x^{1/x}$: $\ln f(x)=\frac{\ln x}x$, so $\dfrac{d}{dx}\ln f(x)=\dfrac{1-\ln x}{x^2}$, negative once $x>e\approx2.718$. So $n^{1/n}$ is decreasing for $n\ge3$ (checking directly: $b_2=\sqrt2\approx1.414<b_3=\sqrt[3]3\approx1.442$, then decreasing thereafter).
Step 3 · Boundedness and conclusion
Since $\lim_{n\to\infty}n^{1/n}=1$ (via $\ln n/n\to0$) and $b_n$ decreases from $b_3=\sqrt[3]3$ down to this limit, $1\le b_n\le\sqrt[3]3<1.5$ for all $n\ge2$: $b_n$ is bounded. With $\sum a_n$ convergent and $b_n$ eventually monotone and bounded, all conditions of Abel's test hold, so the series converges.
Summary Table
| Quantity | Value |
|---|---|
| $\sum(-1)^n/\ln n$ | Converges (Leibniz's test) |
| $b_n=n^{1/n}$ | Decreasing for $n\ge3$, $1\le b_n\le\sqrt[3]3$ |
| Conclusion | Converges (Abel's test) |
Additional Notes
- The limit $n^{1/n}\to1$ and the threshold $n>e$ for monotonicity are the same standard facts established in Problem 70's computation of $\sqrt[n]n\to1$ via logarithms and L'Hôpital's rule — here reused as the bounded multiplier sequence rather than the main object of study.
- This is the third Abel's-test problem in a row (118, 119, 120) built from the same recurring convergent alternating series $\sum(-1)^n/\ln n$, each multiplied by a different bounded, eventually-monotone sequence — showcasing how Abel's test lets a single "core" convergent series generate an entire family of new convergent series.
- Since $b_n>1$ for every $n$ (never touching $0$), this multiplier does not by itself drive the terms of the product to zero the way $\frac1{\ln n}$ alone would; convergence here rests entirely on the oscillation and Leibniz-test convergence of $a_n$, with $b_n$ merely required not to destroy that convergence — exactly the role Abel's test formalizes.
Result
The series converges: