1
$\begingroup$

I'm trying to figure out whether the following series converges absolutely or conditionally or whether it diverges. I am stuck on the following one that involved both sin and cosine: $$ \sum_{n=1}^{\infty} \frac{\sin(n)}{n^2+\cos(n)} $$ Any guidance would be greatly appreciated however :)

$\endgroup$
  • $\begingroup$ Perhaps you can check whether it converges absolutely or not. $\endgroup$ – MathNewbie Jun 11 '15 at 23:07
  • $\begingroup$ Also, as a friendly psa, you can check out the link here to learn how to properly type mathematics on this site to make your equations readable. $\endgroup$ – JMoravitz Jun 11 '15 at 23:11
  • $\begingroup$ That's what I'm asked to do but I'm confused as how to start :) $\endgroup$ – Chloe Jun 11 '15 at 23:11
  • $\begingroup$ Hint: Try and use the fact that $-1\leq\sin(n)\leq 1$. Similar for $\cos(n)$. $\endgroup$ – OnceUponACrinoid Jun 11 '15 at 23:12
  • 1
    $\begingroup$ You can use Limit Comparison with $\sum_1^\infty \frac{1}{n^2}$. Or else use Comparison with $\sum_2^\infty \frac{1}{n^2/2}$. $\endgroup$ – André Nicolas Jun 11 '15 at 23:18
2
$\begingroup$

HINT:

$$\left|\frac{\sin n}{n^2+\cos n}\right|\le \frac{1}{n^2-1}$$

for $n>1$.

$\endgroup$
  • $\begingroup$ Technically you also have to verify $\cos 1 \neq -1$ even if you accept $\cos n \geq -1$, otherwise the series might blow up with the first term and not be defined. $\endgroup$ – user2566092 Jun 11 '15 at 23:21
  • $\begingroup$ @user2566092 this didn't claim to be a full solution, and checking that $\cos 1\neq -1$ is a trivial matter. $\endgroup$ – JMoravitz Jun 11 '15 at 23:23
  • $\begingroup$ I have used the following: the sum from {n=1} to infty of (sin(n))/(n^2 + cos(n)) < (1)/(n^2 -1) < (1)/{n^2} where by the p series the test converges so the series therefore converges absolutely?? $\endgroup$ – Chloe Jun 11 '15 at 23:29
  • $\begingroup$ $\frac{1}{n^2-1} < \frac{1}{n^2}$ is not necessarily true. But you can invoke the limit comparison test between the two series $\sum \frac{1}{n^2-1}$ and $\sum \frac{1}{n^2}$. The p-series test and limit comparison test tells us that $\sum \frac{1}{n^2-1}$ must converge. $\endgroup$ – MathNewbie Jun 11 '15 at 23:31
  • $\begingroup$ So as long as I state that by the comparison test these series hold, the p-series test tells us this so the original series must converge? $\endgroup$ – Chloe Jun 11 '15 at 23:35

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.