# Find $\lim_{n \to \infty} \left(n - \sum_{k=1} ^{n} \cos \frac{\sqrt{k}}{n} \right)$

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.

• Hint: $|1-\cos(x)-x^2/2| \leq x^4/24$ if $|x| \leq 1$. – Mindlack Jan 18 at 17:47

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 $$\frac14$$.

You have $$u_n = n - \sum_{k=1} ^{n} \cos \frac{\sqrt{k}}{n} = \sum_{k=1} ^{n}\left(1-\cos \frac{\sqrt{k}}{n}\right)$$

And as $$\sum_{k=1} ^{n} k = \frac{n(n+1)}{2}$$:

$$u_n - \frac{n+1}{4n} = \sum_{k=1} ^{n}\left(1-\cos \frac{\sqrt{k}}{n}-\frac{k}{2n^2}\right )$$

And using Taylor's theorem, you get for all $$x \in \mathbb R$$

$$\left\vert 1-\cos(x)-x^2/2 \right\vert \leq x^4/24.$$

Therefore $$\left\vert u_n - \frac{n+1}{4n} \right\vert \le \sum_{k=1} ^{n}\left\vert1-\cos \frac{\sqrt{k}}{n}-\frac{k}{4n^2}\right\vert \le \frac{1}{24n^4}\sum_{k=1} ^{n} k^2 \tag{1}$$

As $$\sum_{k=1} ^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$ the RHS of inequality $$(1)$$ converges to $$0$$.

Finaly $$(u_n)$$ converges to $$1/4$$ as $$\lim\limits_{n \to \infty} \frac{n+1}{4n} =1/4$$.