Using Abel's test, prove the convergence of: sum from n=1
Using Abel's test, prove the convergence of: $$ \sum_{n=1}^{\infty}(-1)^n\left(1+\frac{1}{n}\right)^n \tan \frac{1}{n}. $$
Explanation of Method
Abel's test needs a convergent series $\sum a_n$ multiplied by a bounded monotone sequence $b_n$. Grouping $a_n=(-1)^n\tan\frac1n$ (which converges by Leibniz's test, exactly as in Problem 107) and $b_n=\left(1+\frac1n\right)^n$ (which increases monotonically to $e$, hence is bounded) fits this template perfectly — the key insight is choosing the factorization so both pieces satisfy their respective requirements.
- Show $\sum(-1)^n\tan\frac1n$ converges via Leibniz's test.
- Show $b_n=\left(1+\frac1n\right)^n$ is monotonically increasing.
- Show $b_n$ is bounded (converging to $e$), and apply Abel's test.
Step 1 · Convergence of $\sum(-1)^n\tan\frac1n$
As in Problem 107: $\tan\frac1n>0$, decreasing (since $f(x)=\tan\frac1x$ has $f'(x)=-\sec^2(1/x)/x^2<0$), and $\to0$. By Leibniz's test, $\sum_{n=1}^{\infty}(-1)^n\tan\frac1n$ converges.
Step 2 · Monotonicity of $b_n=\left(1+\frac1n\right)^n$
With $f(x)=\left(1+\frac1x\right)^x$, differentiating $\ln f(x)=x\ln\left(1+\frac1x\right)$ gives, after simplification, $\frac{d}{dx}\ln f(x)=\dfrac{x(x-1)}{2x^2(x^2+x)}\cdot(\text{positive factor})>0$ for $x>1$ (using the expansion $\ln(1+1/x)\sim1/x-1/(2x^2)$). So $b_n$ is monotonically increasing.
Step 3 · Boundedness and conclusion
Since $b_n\uparrow e$ (the classical limit), $2=b_1\le b_n<e<3$ for all $n\ge1$: $b_n$ is bounded. All three conditions of Abel's test hold (convergent $\sum a_n$, monotone $b_n$, bounded $b_n$), so the series converges.
Summary Table
| Quantity | Value |
|---|---|
| $\sum(-1)^n\tan\frac1n$ | Converges (Leibniz's test) |
| $b_n=(1+1/n)^n$ | Increasing, $2\le b_n<e<3$ |
| Conclusion | Converges (Abel's test) |
Additional Notes
- Abel's test is often described as "Dirichlet's test with the roles swapped and the boundedness requirement relaxed to allow $\sum a_n$ itself to converge" — this problem is a clean illustration: unlike Dirichlet's test, here $\sum a_n$ must genuinely converge on its own (which it does, via Leibniz), while $b_n$ need only be monotone and bounded, not tend to zero.
- The monotone-increasing sequence $\left(1+\frac1n\right)^n\uparrow e$ is one of the most famous bounded monotone sequences in analysis, and its role here (as the "multiplier" sequence in Abel's test) is a nice application distinct from its more usual appearance as the defining limit of $e$ itself.
- Since $b_n$ converges to $e$ rather than to $0$, the resulting series' sum is genuinely different from (roughly $e$ times) the sum of $\sum(-1)^n\tan\frac1n$ alone — Abel's test guarantees convergence but not this simple multiplicative relationship, since $b_n$ varies with $n$.
Result
The series converges by Abel's test: