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...


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?

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    $\begingroup$ That sandwich is delish ;-) +1 $\endgroup$ – Vizag Apr 20 '19 at 19:17
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    $\begingroup$ I don't understand why $\cos(\frac{2}{n}) <= \cos(\frac{n+k}{n^2})$ ? $\endgroup$ – PhysicsPrincess Apr 20 '19 at 20:34
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    $\begingroup$ 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

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