Find $\lim_{n \to \infty} \left(n – \sum_{k=1} ^{n} \cos \frac{\sqrt{k}}{n} \right)$
My Attempt:
$\forall x: \ |\cos(x)|\leq 1$, Therefore:
$$\lim_{n \to \infty} \left(n – \sum_{k=1} ^{n} \cos \frac{\sqrt{k}}{n} \right) \leq \lim_{n \to \infty} \left(n – \left| \sum_{k=1} ^{n} \cos \frac{\sqrt{k}}{n} \right| \right) \leq \lim_{n \to \infty} \left( \sum_{k=1}^{n} |\cos \frac{\sqrt{k}}{n} | \right) \leq \lim_{n \to \infty} \left(n – \sum_{k=1} ^{n} k \right) = 0$$
but I can't find a way to bound the limit such that I could prove that:
$$0 \leq \lim_{n \to \infty} \left(n – \sum_{k=1} ^{n} \cos \frac{\sqrt{k}}{n} \right) \leq 0$$
which would finish the proof.
Best Answer
Using the estimate $\cos(x)=1-\frac12\,x^2+O(x^4)$, we get $$ \lim_{n\to\infty} \left(n-\sum_{k=1}^{n} \cos\frac{\sqrt{k}}{n} \right) = \lim_{n\to\infty} \sum_{k=1}^n \left( 1-\cos\frac{\sqrt k}n\right) = \lim_{n\to\infty} \sum_{k=1}^n \left( \frac k{2n^2} + O\Big(\frac{k^2}{n^4}\Big) \right). $$ The sum splits into $$ \sum_{k=1}^n \frac k{2n^2} = \frac1{2n^2} \frac{n(n+1)}2 $$ converging to $\frac14$, and the remainder term bounded by $$ \frac1{n^4}\frac{n(n+1)(2n+1)}6 , $$ which converges to $0$. Therefore, your original limit is $\frac14$.