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?

  • $\begingroup$ 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. $\endgroup$ – Varun Vejalla Sep 2 '20 at 4:13
  • $\begingroup$ @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. $\endgroup$ – Brian Moehring Sep 2 '20 at 4:19
  • $\begingroup$ 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 $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}\sin k}{2k} = \frac{1}{4}$. $\endgroup$ – João Júnior Sep 2 '20 at 4:21
  • $\begingroup$ 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. $\endgroup$ – João Júnior Sep 2 '20 at 4:26
  • 1
    $\begingroup$ 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. $\endgroup$ – Brian Moehring Sep 2 '20 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.

  • 1
    $\begingroup$ Of course the other part $\sum_{k=1}^\infty \frac{(-1)^k}{k}$ converges by AST. $\endgroup$ – RRL Sep 2 '20 at 4:35
  • $\begingroup$ $\cos (\pi k) \sin k = \sin(\pi k + k)$? $\endgroup$ – João Júnior Sep 2 '20 at 4:41
  • 1
    $\begingroup$ $\sin(a+b) = \sin a \cos b + \sin b \cos a$ and $\sin \pi k = 0$ $\endgroup$ – RRL Sep 2 '20 at 4:50
  • $\begingroup$ Sorry, I'm sleepy, underperforming. $\endgroup$ – João Júnior Sep 2 '20 at 4:52
  • $\begingroup$ @RRL Hi, could I ask your assistance here, please? $\endgroup$ – Antonio Maria Di Mauro Sep 2 '20 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.