Does the following series converge? If so, does it converge absolutely?

$$\sum_{n=1}^{\infty} \tan\left(\frac{(-1)^n}{n}\right)$$

My attempt:

$$\sum_{n=1}^{\infty} \tan\left(\frac{(-1)^n}{n}\right) = \sum_{n=1}^{\infty} (-1)^n\tan\left(\frac{1}{n}\right)$$

and because $\tan$ is an increasing function on $(0,1]$, we have:

$$\tan\left(\frac{1}{n+1}\right) < \tan\left(\frac{1}{n}\right)$$

Also, $\tan\left(\frac{1}{n}\right) \to 0$. Hence, by Leibniz criterion the series converges.

To examine absolute convergence, we must examine the series:

$$\sum_{n=1}^{\infty} \tan\left(\frac{1}{n}\right)$$

Notice that $$\frac{\tan\left(\frac{1}{n}\right)}{\frac{1}{n}} \to 1$$

Hence, there is $n_0$ such that $\forall n \geq n_0: $

$$\left|\frac{\tan\left(\frac{1}{n}\right)}{\frac{1}{n}}-1\right| < \frac{1}{2}$$

Hence, for $n \geq n_0: $

$$\frac{\tan\left(\frac{1}{n}\right)}{\frac{1}{n}} \geq \frac{1}{2}$$

Thus, $\tan\left(\frac{1}{n}\right) \geq \frac{1}{2n}$

and by comparing with the harmonic series, it follows that the series diverges, and hence the series is not absolutely convergent. Is this correct? Are there easier approaches? Thanks in advance.

  • $\begingroup$ The series converges trivially by the alternating series test, but does not converge absolutely by comparison with the harmonic series. $\endgroup$ – Frpzzd Dec 25 '17 at 22:12
  • $\begingroup$ Yes, it is correct. $\endgroup$ – José Carlos Santos Dec 25 '17 at 22:15
  • $\begingroup$ The limit conparison test (ratio approaching $1$) should be enough to tell you it doesn't converge absolutely $\endgroup$ – Dylan Dec 25 '17 at 22:44

Your Answer

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