Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty}(-1)^n \tan \frac{1}{n}. $$
Explanation of Method
Since $\tan\frac1n\sim\frac1n$ for large $n$, the absolute series $\sum\tan\frac1n$ has the same convergence behavior as the harmonic series by the limit comparison test — hence divergence, so no absolute convergence. But $\tan\frac1n$ is positive, decreasing (as the composition of a decreasing $\frac1n$ with the increasing $\tan$ on $(0,\pi/2)$), and tends to $0$, so Leibniz's test applies directly, giving conditional convergence.
- Compare $\tan\frac1n$ with $\frac1n$ via the limit comparison test, showing $\sum\tan\frac1n$ diverges.
- Verify the three Leibniz conditions for $b_n=\tan\frac1n$.
- Conclude conditional (not absolute) convergence.
Step 1 · Test for absolute convergence
With $v_n=\frac1n$: $\displaystyle\lim_{n\to\infty}\frac{\tan(1/n)}{1/n} = \lim_{u\to0}\frac{\tan u}u = 1$, a finite nonzero constant. Since $\sum\frac1n$ diverges, by the limit comparison test $\sum\tan\frac1n$ diverges too — the series does not converge absolutely.
Step 2 · Verify Leibniz's conditions for $b_n=\tan\frac1n$
Since $0<\frac1n<\frac\pi2$, $b_n=\tan\frac1n>0$. The function $f(x)=\tan\frac1x$ has $f'(x)=-\dfrac{\sec^2(1/x)}{x^2}<0$, so $b_n$ is strictly decreasing. And $\lim_{n\to\infty}\tan\frac1n=\tan0=0$.
Step 3 · Apply Leibniz's test
All three conditions hold, so by Leibniz's test the series converges. Combined with Step 1, the series converges conditionally.
Summary Table
| Test | Result |
|---|---|
| $\sum\tan\frac1n$ (absolute series) | Diverges (limit comparison with $\sum1/n$) |
| Leibniz's test | All conditions satisfied |
| Conclusion | Conditionally convergent |
Additional Notes
- The asymptotic equivalence $\tan\frac1n\sim\frac1n$ places this problem in exactly the same convergence class as Problem 103 with $\alpha=1$: both are conditionally convergent alternating series whose underlying positive series matches the divergent harmonic series.
- Monotonicity here follows purely from composing a decreasing function ($1/x$) with an increasing one ($\tan$ on $(0,\pi/2)$), a general principle: decreasing $\circ$ increasing $=$ decreasing, useful whenever checking Leibniz's monotonicity condition for a composite expression.
- This series and Problem 108's $\sum(-1)^n/\ln n$ share the same qualitative structure (positive, decreasing, $\to0$, but with divergent absolute series) — both are standard illustrations of conditional convergence beyond the basic alternating-harmonic-series example.
Result
The series converges conditionally but not absolutely: