# The alternating series $\sum_{k=1}^{\infty} \frac{(-1)^k(2 - \sin k)}{2k}$ seems to be convergent, but Leibniz criterion does not apply

I was looking for an example of convergent, alternating series $$\sum_{k=1}^{\infty}(-1)^kb_k$$ such that $$\{b_k\}_{k=1}^{\infty}$$ is not eventually monotone, so that Leibiniz criterion could not be applied. Preferably, one whose convergence is conditional (not absolute). So, I thought of $$\sum_{k=1}^{\infty} \frac{(-1)^k(2 - \sin k)}{2k}$$. WolframAlpha says that this series converges, and it is clearly not absolutely convergent. But I am trying to prove its convergence, and I haven't been successful so far. Does anyone have any ideas?

• The exact value of the series is $\frac{1}{4}-\ln 2$. Splitting the sum as $\frac{(-1)^k}{k}$ and $\frac{(-1)^k \sin k}{2k}$ may help. Or realizing that $2-\sin k \le 3$ and then using the alternating series test. I don't know if either of these methods is mathematically sound though. Commented Sep 2, 2020 at 4:13
• @Varun Splitting the sum is justified by showing one of the two resulting series is finite, which is elementary for the first one, so that's fine. Bounding $2-\sin(x) \leq 3$ doesn't help, though. Commented Sep 2, 2020 at 4:19
• If $\sum_{k=1}^{\infty}\frac{(-1)^k(2 - \sin k)}{2k} = \frac{1}{4} - \ln 2$, and $\sum_{k=1}^{\infty}\frac{(-1)^k}{k} = -\ln 2$, then $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}\sin k}{2k} = \frac{1}{4}$. Commented Sep 2, 2020 at 4:21
• The problem now is: how to prove that $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}\sin k}{2k}$ converges? Equivalently, how to prove that $\sum_{k=1}^{\infty}\frac{(-1)^{k}\sin k}{k}$ converges? This is another alternating series $\sum_{k=1}^{\infty} (-1)^kb_k$ such that $\{b_k\}_{k=1}^{\infty}$ is not eventually monotone, so Leibniz criterion does no apply. Commented Sep 2, 2020 at 4:26
• Note that $\sum \frac{\sin k}{k}$ doesn't converge absolutely (i assume you mistakenly thought $\sin k \geq 0$). Abel's test tells us that $\sum \frac{1}{k} z^k$ converges for all $|z|\leq 1, z\neq 1$, and setting $z=-e^i$ gives us the series we want. Commented Sep 2, 2020 at 4:43

Note that

$$\sum_{k=1}^n\frac{(-1)^k \sin k}{2k} = \sum_{k=1}^n\frac{\cos (\pi k) \sin k}{2k} = \sum_{k=1}^n\frac{\sin ((\pi +1)k)}{2k}$$

and the right hand side series converges by Dirichlet's test.

• Of course the other part $\sum_{k=1}^\infty \frac{(-1)^k}{k}$ converges by AST.
– RRL
Commented Sep 2, 2020 at 4:35
• $\cos (\pi k) \sin k = \sin(\pi k + k)$? Commented Sep 2, 2020 at 4:41
• $\sin(a+b) = \sin a \cos b + \sin b \cos a$ and $\sin \pi k = 0$
– RRL
Commented Sep 2, 2020 at 4:50
• Sorry, I'm sleepy, underperforming. Commented Sep 2, 2020 at 4:52