Convergence of $\sum (-1)^n \sin^2(n)/n$ 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.
 A: 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.
A: Note that
$$\sum_{n=1}^{\infty} \frac{(-1)^n\sin^2(n)}{n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{2n} - \sum_{n=1}^{\infty} \frac{(-1)^n\cos(2n)}{2n}$$
which converges since


*

*$\sum \frac{(-1)^n}{2n}$ converges

*$\sum \frac{(-1)^n\cos(2n)}{2n}$ converges 


indeed by Abel transformation and Lagrange's trigonometric identities, let 
$$a_n=\frac{(-1)^n}{2n} \quad b_n=\cos(2n)=B_{n}-B_{n-1}\quad B_n=\sum_{k=1}^{n} \cos (2k)= -\frac12+\frac{\sin(2n+1)}{2\sin 1}$$
$$S_N=\sum_{n=1}^{N}a_nb_n=\sum_{n=1}^{N} \frac{\cos(2n)}{2n}
=\frac{(-1)^N}{2N}\left(-\frac12+\frac{\sin(2N+1)}{2\sin 1}\right)-\sum_{n=1}^{N-1} \left[ \left(-\frac12+\frac{\sin(2n+1)}{2\sin 1}\right)\left(-\frac{(-1)^n}{2n+2}-\frac{(-1)^n}{2n}\right)\right]=$$
$$=\frac{(-1)^N}{2N}\left(-\frac12+\frac{\sin(2N+1)}{2\sin 1}\right)+\sum_{n=1}^{N-1} \frac{(-1)^n(2n+1)}{2n(n+1)}-\sum_{n=1}^{N-1} \frac{(-1)^n\sin (2n+1)}{4n(n+1)\sin 1}$$
and
$$\sum_{n=1}^{\infty} \frac{(-1)^n\sin (2n+1)}{4n(n+1)\sin 1}$$
converges absolutely by comparison with $\sum \frac{1}{n^2}$.
A: \begin{align*}
\sum(-1)^{n}\dfrac{\sin^{2}n}{n}=\dfrac{1}{2}\sum(-1)^{n}\left(\dfrac{1}{n}-\dfrac{\cos 2n}{n}\right),
\end{align*}
but
\begin{align*}
\sum(-1)^{n}\dfrac{\cos 2n}{n}
\end{align*}
converges because of the Dirichlet test:
\begin{align*}
\sum(-1)^{n}\dfrac{\cos 2n}{n}=\sum\left((-1)^{n}\cos 2n\right)\cdot\dfrac{1}{n},
\end{align*}
and that 
\begin{align*}
\left|\sum_{k=1}^{n}(-1)^{k}\cos(2k)\right|&=\left|\text{Re}\sum_{k=1}^{n}(-1)^{k}e^{2ki}\right|\leq\left|\sum_{k=1}^{n}(-1)^{k}e^{2ki}\right|
\end{align*}
is uniformly bounded on $n$.
