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I'm trying to solve one of the tasks in the Algorithm Design Manual book from Steven Skiena. The goal is to place the functions into increasing asymptotic order.


$f_2(n)=\sqrt{n} \log{n}$,

$f_3(n)=n \sqrt{\log{n}}$,


With $f_2(n)$, $f_3(n)$, $f_4(n)$ it's pretty simple. But I don't understand how to resolve this function $f_1(n)=\sum_{i=1}^n\sqrt{i}$.

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up vote 1 down vote accepted

Hint: $$\int_0^n\sqrt{x}dx\le\sum_{i=1}^n\sqrt{i}\le\int_0^n\sqrt{x+1}dx$$

Edit: We know that $x \le \lceil x\rceil\le x+1$, so $$\int_0^n\sqrt{x}dx\le\int_0^n\sqrt{\lceil x\rceil}dx\le\int_0^n\sqrt{x+1}dx$$ but for $x\in (n-1,n]$ we have $\sqrt{\lceil x\rceil}=\sqrt{n}$, so we can write $$\int_0^n\sqrt{\lceil x\rceil}dx=\sum_{i=1}^n \int_{i-1}^i\sqrt{i}dx=\sum_{i=1}^n\sqrt{i}$$

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Thank you. Can you maybe point me to the proof of this inequality? – Igor Konoplyanko Jan 10 '13 at 11:50

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