Using Abel's test, prove the convergence of: sum from n=2
Using Abel's test, prove the convergence of: $$ \sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n} \frac{\sqrt{n}}{n+a}, \qquad a>0. $$
Explanation of Method
Factoring the term as $a_n=\frac{(-1)^n}{\ln n}$ (which converges by Leibniz's test, exactly as in Problem 108) times $b_n=\frac{\sqrt n}{n+a}$ sets up Abel's test cleanly. The function $\frac{\sqrt x}{x+a}$ turns out to be decreasing once $x>a$, and it is always bounded between $0$ and $\frac1{\sqrt2}$ — so $b_n$ is eventually monotone and uniformly bounded, exactly what Abel's test requires.
- Show $\sum(-1)^n/\ln n$ converges via Leibniz's test.
- Show $b_n=\sqrt n/(n+a)$ is eventually decreasing (for $n>a$).
- Show $b_n$ is bounded, and apply Abel's test.
Step 1 · Convergence of $\sum(-1)^n/\ln n$
As in Problem 108: $\frac1{\ln n}>0$ for $n\ge2$, decreasing, and $\to0$. By Leibniz's test, $\sum_{n=2}^{\infty}\dfrac{(-1)^n}{\ln n}$ converges.
Step 2 · Eventual monotonicity of $b_n=\sqrt n/(n+a)$
With $f(x)=\dfrac{\sqrt x}{x+a}$: $f'(x)=\dfrac{a-x}{2\sqrt x(x+a)^2}$, which is negative for $x>a$. So $b_n$ is decreasing for all $n>a$; splitting off the finitely many terms with $n\le a$ doesn't affect convergence.
Step 3 · Boundedness and conclusion
For all $n\ge2$: $0<b_n=\dfrac{\sqrt n}{n+a}<\dfrac{\sqrt n}n=\dfrac1{\sqrt n}\le\dfrac1{\sqrt2}$, so $b_n$ is bounded. With $\sum a_n$ convergent (Step 1) and $b_n$ eventually monotone and bounded (Steps 2–3), all conditions of Abel's test hold, so the series converges for every $a>0$.
Summary Table
| Quantity | Value |
|---|---|
| $\sum(-1)^n/\ln n$ | Converges (Leibniz's test) |
| $b_n=\sqrt n/(n+a)$ | Eventually decreasing, $0<b_n\le1/\sqrt2$ |
| Conclusion | Converges for all $a>0$ (Abel's test) |
Additional Notes
- This problem is essentially Problem 108's series (which supplies the convergent $\sum a_n$ half) multiplied by a bounded, eventually-decreasing rational factor — a natural way to generate new Abel's-test examples from previously established Leibniz-test results.
- The eventual monotonicity threshold $n>a$ is analogous to the threshold seen in Problem 116's split at $n=8$: both problems require isolating a finite initial segment before the desired monotonicity kicks in, a routine but essential step whenever the multiplier sequence isn't monotone from the very start.
- Since $b_n\to0$ as well (not just bounded), this series would also satisfy Dirichlet's test if one instead grouped $a_n=(-1)^n$ (with bounded partial sums) against the full decreasing-to-zero product $\frac1{\ln n}\cdot\frac{\sqrt n}{n+a}$ — Abel's and Dirichlet's tests often overlap in their applicability for series like this one.
Result
The series converges for every $a>0$: