$(\sum_{k\in[\sqrt{n},n-\sqrt{n}]}1)$ vanishes in $\frac{1}{n}(\sum_{k\in[\sqrt{n},n-\sqrt{n}]}1)(M+|\alpha|)$

Question

The following derivation took place in another thread.

$\Bigl|\frac{1}{n}\sum\limits_{k\in[\sqrt{n},n-\sqrt{n}]}a_kb_{n-k}-\alpha\beta\,\Bigr|\le\frac{1}{n}\,\Bigl(\sum\limits_{k\in[\sqrt{n},n-\sqrt{n}]}1\Bigr)(M+|\alpha|)\,\epsilon\le(M+|\alpha|)\,\epsilon$

However I am having a hard time understanding how $\sum\limits_{k\in[\sqrt{n},n-\sqrt{n}]}1$ vanishes. I thought that $n-\sqrt{n}-\sqrt{n}=n-2\sqrt{n}=1$, but I have no idea if this is the way and if this last equality is true.

Question:

Could someone explain me how $\sum\limits_{k\in[\sqrt{n},n-\sqrt{n}]}1$ vanishes?

Thanks in advance!

To prevent potential doubts here is the link to the other thread $\frac{a_0b_n+a_1b_{n-1}+...+a_nb_0}{n}$ converges to $\alpha\beta$

Answer 1

$$\frac1n\big(\sum_{k\in [\sqrt{n},n-\sqrt{n}]}1\big)\le \frac1n\big(\sum_{k=1}^n1\big)=\frac1n\cdot n=1,$$ so $$ \frac1n\big(\sum_{k\in [\sqrt{n},n-\sqrt{n}]}1\big)(M+|\alpha|)\epsilon \le (M+|\alpha|)\epsilon $$