Convergence of $\sum (-1)^n \sin^2(n)/n$

Question

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.

Answer 1

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).

The theorem is called Dirichlet's test. It states:

• Let $a_{0}>a_{1}>a_{2}>\ldots$ be a decreasing real sequence tending to $0$,
• $z \in \mathbb{C}$ 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.

Answer 2

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 $\sum \frac{1}{n^2}$.

Answer 3

\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 $n$.