1
$\begingroup$

The question is to determine whether the following series is convergent or divergent:

$$ \sum_{n = 1}^{\infty}\left(-1\right)^{n - 1}\sin\left(1 \over n\right) $$

I used the alternate series test in order to test for convergence. $\left\vert a_{n}\right\vert = \sin\left(1/n\right)$ and $\lim_{n \to \infty}\sin\left(1/n\right)$ does not exist. However, I'm not sure if the series meets the second precondition for this test $\left(~a_{n + 1} < a_{n}~\right)$, so I'm a bit stuck on other methods to test for convergence.

I also looked up the series on WolframAlpha and there were some inconsistent results. It gave an imaginary sum for the series, but the partial sums looked like they were converging at some value. Any help would be appreciated !.

$\endgroup$
  • $\begingroup$ Actually, $\lim\limits_{n\to\infty}\sin(1/n)=0$, so the alternating test does apply. Since $\frac d{dx}\sin(x)=\cos(x)>0$ for $x\in[0,1]$, it follows that $\sin(a)>\sin(b)$ when $a>b$ and $a,b\in[0,1]$, thus $|a_{n+1}|<|a_n|$. $\endgroup$ – Simply Beautiful Art Mar 28 '17 at 16:10
1
$\begingroup$

There are only a few conditions that need to be checked in order to apply the alternating series test for convergence.

  1. Is your series alternating?

    Yes, and this is easy to see.

  2. Is $\sin(\frac{1}{n+1}) < \sin(\frac{1}{n})$?

    Yes. Thinking geometrically, is is clear that if $0 \leq \theta_1 < \theta_2 \leq \pi/2$, then $\sin(\theta_1) < \sin(\theta_2)$. Indeed, this is simply stating that on the unit circle in the first quadrant, bigger angles lead to higher points.

  3. Is $\lim_{n \to \infty} \sin(\frac{1}{n}) = 0$?

    Yes. In particular, the sine function is continuous, and so $$ \lim_{n \to \infty} \sin ( \frac{1}{n} ) = \sin( \lim_{n \to \infty} \frac{1}{n} ) = \sin (0) = 0.$$ Thus the limit of the summands is zero.

Putting these three conditions together, it follows that the series converges by the alternating series test.

| cite | improve this answer | |
$\endgroup$

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.