5
$\begingroup$

Hi I could really use some help with this homework question.

$$\lim_{n\to\infty} \frac{1}{n} \sum\limits_{k=1}^{n} \cos{\left(\frac{n+k}{n^2}\right)}$$

I have no idea how to solve it (we haven't learned sentences about series yet...)

My attempts:

I plotted the function and it looks like the limit is $0$.

I tried to find a bound to this Cosinus series without success (I think it might be unbounded but I'm not sure).

I also tried to use the identity of angle addition: $$\cos\left(\frac{n+k}{n^2}\right) = \cos\left(\frac{1}{n} + \frac{k}{n^2}\right) = \cos\left(\frac{1}{n}\right)\cos\left(\frac{k}{n^2}\right) - \sin\left(\frac{1}{n}\right)\sin\left(\frac{k}{n^2}\right)$$ but it leads to nothing...

How do I even approach a question like this? I can't seen to be able to bound it trivially or use arithmetic rules...

$\endgroup$
10
$\begingroup$

Hint: If $1 \le k \le n$, then we have $$\cos\left(\dfrac{2}{n}\right) \le \cos\left(\dfrac{n+k}{n^2}\right) \le 1.$$ Hence, $$\dfrac{1}{n}\sum_{k = 1}^{n}\cos\left(\dfrac{2}{n}\right) \le \dfrac{1}{n}\sum_{k = 1}^{n}\cos\left(\dfrac{n+k}{n^2}\right) \le \dfrac{1}{n}\sum_{k = 1}^{n}1,$$ i.e. $$\cos\left(\dfrac{2}{n}\right) \le \dfrac{1}{n}\sum_{k = 1}^{n}\cos\left(\dfrac{n+k}{n^2}\right) \le 1.$$ Can you figure out the limit from here?

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ That sandwich is delish ;-) +1 $\endgroup$ – Vizag Apr 20 '19 at 19:17
  • 2
    $\begingroup$ I don't understand why $\cos(\frac{2}{n}) <= \cos(\frac{n+k}{n^2})$ ? $\endgroup$ – PhysicsPrincess Apr 20 '19 at 20:34
  • 3
    $\begingroup$ The function $\cos x$ is decreasing on $[0,\pi]$. Then since $0 \le \tfrac{n+k}{n^2} \le \tfrac{2}{n} \le \pi$, we have $\cos\left(\tfrac{2}{n}\right) \le \cos\left(\tfrac{n+k}{n^2}\right)$. $\endgroup$ – JimmyK4542 Apr 20 '19 at 21:13

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.