0
$\begingroup$

Consider whether the following series converges or diverges $$\sum_{n=1}^\infty\frac{(-1)^n\sin\frac{n\pi}{3n+1}}{\sqrt{n+3}}$$

I have tried Leibniz's test but failed to show that $\sin\frac{n\pi}{3n+1} / \sqrt{n+3}$ is monotone, which I graphically checked it is. I have managed to show that the series diverges absolutely, therefore I've shown nothing. Also I have tried using Cauchy's criterion, but failed to arrive at any useful result.

$\endgroup$
2
$\begingroup$

You can use Abel's test:

  • Set $a_n := \frac{(-1)^n}{\sqrt{n+3}}$ and $b_n := \sin \frac{n\pi}{3n+1}$
  • $\sum_{n=1}^{\infty}a_n$ is convergent
  • Note that $b_n = \sin \frac{\pi}{3}\left(1-\frac 1{3n+1} \right)$, so $b_n$ is increasing and bounded.

Now, invoking Abel's test (which uses summation by parts), we can conclude that $\sum_{n=1}^{\infty}a_nb_n = \sum_{n=1}^\infty\frac{(-1)^n\sin\frac{n\pi}{3n+1}}{\sqrt{n+3}}$ is convergent.

$\endgroup$
2
$\begingroup$

Observe that

$$\sin\frac{(n+1)\pi}{3n+4}\le\sin\frac{n\pi}{3n+1}\iff\sin\frac{(n+1)\pi}{3n+4}-\sin\frac{n\pi}{3n+1}\le0\stackrel{\text{trig. identities}}\iff$$

$$\iff2\sin\frac\pi{(3n+1)(3n+4)}\cdot\cos\frac{6n^2+8n+1}{(3n+1)(3n+4)}\pi\le0$$

But in the last expression above the angle of the sine is in the first quadrant and thus its sine is positive, whereas the angle of the the cosine is between $\;\pi/2\;$ and $\pi\;$ for any values of $\;n\;$ (why? This is a nice exercise in quadratic functions in natural numbers...) so its cosine is negative and thus your sequence is the product of two decreasing positive sequences....

$\endgroup$

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.