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

How can I calculate the following limit:

$$\lim \limits_{n \to \infty}(n-\sum_{k=1}^n\cos{\frac{\sqrt{k}}{n})}$$

A hint or direction would be appreciated (please not a solution for now, I would post mine once I get it).

I have tried to use 3rd order of taylor but I couldn't get through with the algebra.

Thank you

• Hint : $1-\cos(\frac{\sqrt{k}}{n})\sim \frac{k}{n^2}$ ... (since $k\leq n$)... What do you think of $\sum_n \frac{k}{n^2}$? – Netchaiev Jan 29 '18 at 12:07

Note that using Taylor series to first term, you get

$$n-\sum_{k=1}^{n}\cos\frac{\sqrt{k}}{n}\approx\sum_{k=1}^{n} \frac{k}{2 n^2} = \frac{1}{2}\int_{0}^{1} xdx = \frac{1}{4}$$

Alternatively you can use $\sum_{k=1}^n k = \frac{n(n+1)}{2}$

• "(please not a solution for now"... But it is a nice use of my hint... – Netchaiev Jan 29 '18 at 12:27
• Sorry did not read!! – King Tut Jan 29 '18 at 12:31
• I have not studied integrals and the use of this notation: $\approx$. Can you explain? – Mister Bister Jan 29 '18 at 12:33
• What does it mean? How did you make that transition? – Mister Bister Jan 29 '18 at 12:41
• Yeah! And Then numerator degree cannot keep up with denominator degree rising – King Tut Jan 29 '18 at 12:56

Hint:

$$n-\sum_{k=1}^{n}\cos\frac{\sqrt{k}}{n}=\sum_{k=1}^{n}\left(1-\cos\frac{\sqrt{k}}{n}\right)=2\sum_{k=1}^{n}\sin^2 \frac{\sqrt{k}}{2n}$$ and $\frac{\sqrt{k}}{n}\in(0,1)$. You may consider that $\sin^2 x = x^2+O(x^4)$ and that $$\sum_{k=1}^{n}\frac{k}{4n^2}=\frac{1}{8}+o(1),\qquad \sum_{k=1}^{n}\frac{k^2}{16n^4}=o(1).$$