# Finding the limit of $\lim_{n\to\infty} \frac{1}{n} \sum\limits_{k=1}^{n} \cos{(\frac{n+k}{n^2})}$

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...

• – lab bhattacharjee Apr 20 '19 at 18:58
• Are you sure about the value of limit? I would expect it to be 1. – user Apr 20 '19 at 19:03
• I am not sure at all.. It looks weird when trying to plot it. – PhysicsPrincess Apr 20 '19 at 19:04

## 1 Answer

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?

• That sandwich is delish ;-) +1 – Vizag Apr 20 '19 at 19:17
• I don't understand why $\cos(\frac{2}{n}) <= \cos(\frac{n+k}{n^2})$ ? – PhysicsPrincess Apr 20 '19 at 20:34
• 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 $\cos\left(\tfrac{2}{n}\right) \le \cos\left(\tfrac{n+k}{n^2}\right)$. – JimmyK4542 Apr 20 '19 at 21:13