The series $\sum (-1)^n \frac{\sin^2(n)}{n}$ is a good example of something that fails the Alternating Series Test since the corresponding positive terms are not monotonic. The notes I took a couple years ago say "It converges (conditionally) via clever trig identities." But now I can't figure out how again.
Calculus – Convergence of $\sum (-1)^n \sin^2(n)/n$
calculusconvergence-divergencesequences-and-series
Best Answer
Writing $\sin^{2}(n) = \frac{1}{2}(1-\cos(2n))$, we are reduced to showing $$\sum (-1)^{n} \frac{\cos(2n)}{n} = \text{Re}\sum \frac{z^{n}}{n}$$ where $z = e^{i(\pi+2)}$. This complex series converges because $\pi+2$ is not an even multiple of $\pi$ (this is a complex generalisation of the alternating series test).
I thought this generalisation went by the name `Abel's lemma', but I can't find it. The theorem is: Let $a_{0}>a_{1}>a_{2}>\ldots$ be a decreasing real sequence tending to $0$, and $z \in \mathbb{C}$ with $|z|=1$ but $z \ne 1$. Then $$\sum_{n=1}^{\infty}a_{n}z^{n}$$ converges. The proof is the same as the alternating series test: $$(1-z)\sum_{n=N}^{\infty}a_{n}z^{n} = a_{N}z^{N} + \sum_{n=N+1}(a_{n+1}-a_{n})z^{n}$$ If $N$ is large enough to imply $a_{n} < \varepsilon$ for all $n>N$, then $$|RHS| < |a_{N}||z|^{N} + \sum_{n=N+1}|a_{n+1}-a_{n}||z|^{n} \\ < \varepsilon - \sum_{n=N+1}(a_{n}-a_{n+1})\\ = \varepsilon - a_{N+1}\\ < 2\varepsilon$$ So the series converges.